AREA THEOREMS [Proofs and Uses] - Questions & Answers
EXERCISE 151. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) In the given figure, BD : DC = 3 : 5, then area of ΔABD : area of ΔACD is :
(i) 5 : 3
(ii) 3 : 5
(iii) 25 : 9
(iv) 9 : 25
Answer: (ii) 3 : 5
Step 1: Triangles ABD and ACD share the same vertex A.
Step 2: Their bases BD and DC lie on the same straight line BC.
Step 3: Therefore, their heights are equal.
Step 4: The ratio of areas of triangles with the same height is equal to the ratio of their bases.
Step 5: Area of ΔABD : Area of ΔACD = BD : DC = 3 : 5.
Step 1: Triangles ABD and ACD share the same vertex A.
Step 2: Their bases BD and DC lie on the same straight line BC.
Step 3: Therefore, their heights are equal.
Step 4: The ratio of areas of triangles with the same height is equal to the ratio of their bases.
Step 5: Area of ΔABD : Area of ΔACD = BD : DC = 3 : 5.
(b) A median of a triangle divides it into two :
(i) triangle of equal areas.
(ii) congruent triangles
(iii) triangles of areas in the ratio 2 : 1.
(iv) right triangles
Answer: (i) triangle of equal areas.
Step 1: A median connects a vertex to the midpoint of the opposite side, dividing the base into two equal segments.
Step 2: The two smaller triangles share the same vertex and have equal bases.
Step 3: Thus, their areas are equal.
Step 1: A median connects a vertex to the midpoint of the opposite side, dividing the base into two equal segments.
Step 2: The two smaller triangles share the same vertex and have equal bases.
Step 3: Thus, their areas are equal.
(c) In the given figure, AB is parallel to DC and AB ≠ DC, the area of ΔAOD is equal to area of triangle :
(i) AOB
(ii) COD
(iii) ACB
(iv) BOC
Answer: (iv) BOC
Step 1: Triangle ABD and Triangle ABC are on the same base AB.
Step 2: They lie between the same parallel lines AB and DC.
Step 3: Therefore, Area(ΔABD) = Area(ΔABC).
Step 4: Subtract the common Area(ΔAOB) from both sides.
Step 5: Area(ΔABD) - Area(ΔAOB) = Area(ΔABC) - Area(ΔAOB).
Step 6: Area(ΔAOD) = Area(ΔBOC).
Step 1: Triangle ABD and Triangle ABC are on the same base AB.
Step 2: They lie between the same parallel lines AB and DC.
Step 3: Therefore, Area(ΔABD) = Area(ΔABC).
Step 4: Subtract the common Area(ΔAOB) from both sides.
Step 5: Area(ΔABD) - Area(ΔAOB) = Area(ΔABC) - Area(ΔAOB).
Step 6: Area(ΔAOD) = Area(ΔBOC).
(d) In the given figure, AF//BE and PQ// RS, FC and ED are perpendiculars to RS. The area of parallelogram ABEF is equal to :
(i) rect. CDEF
(ii) quad. CBEF
(iii) 2 x ΔACF
(iv) 2 x ΔEBD
Answer: (i) rect. CDEF
Step 1: Parallelogram ABEF and rectangle CDEF stand on the same base EF.
Step 2: They lie between the same parallel lines PQ (containing E, F) and RS (containing C, D).
Step 3: According to the area theorem, a parallelogram and rectangle on the same base and between same parallels are equal in area.
Step 4: Therefore, Area of ABEF = Area of rect. CDEF.
Step 1: Parallelogram ABEF and rectangle CDEF stand on the same base EF.
Step 2: They lie between the same parallel lines PQ (containing E, F) and RS (containing C, D).
Step 3: According to the area theorem, a parallelogram and rectangle on the same base and between same parallels are equal in area.
Step 4: Therefore, Area of ABEF = Area of rect. CDEF.
(e) In the given figure, D is mid-point of side BC, the area of triangle BEA is equal to area of triangle :
(i) BED
(ii) CED
(iii) CEA
(iv) ACD
Answer: (iii) CEA
Step 1: In ΔABC, AD is the median because D is the midpoint of BC.
Step 2: Therefore, Area(ΔABD) = Area(ΔACD).
Step 3: In ΔEBC, ED is the median.
Step 4: Therefore, Area(ΔEBD) = Area(ΔECD).
Step 5: Subtract the second equation from the first.
Step 6: Area(ΔABD) - Area(ΔEBD) = Area(ΔACD) - Area(ΔECD).
Step 7: Area(ΔBEA) = Area(ΔCEA).
Step 1: In ΔABC, AD is the median because D is the midpoint of BC.
Step 2: Therefore, Area(ΔABD) = Area(ΔACD).
Step 3: In ΔEBC, ED is the median.
Step 4: Therefore, Area(ΔEBD) = Area(ΔECD).
Step 5: Subtract the second equation from the first.
Step 6: Area(ΔABD) - Area(ΔEBD) = Area(ΔACD) - Area(ΔECD).
Step 7: Area(ΔBEA) = Area(ΔCEA).
2. In the given figure, if area of triangle ADE is 60 cm²; state, giving reason, the area of :
(i) parallelogram ABED;
(ii) rectangle ABCF;
(iii) triangle ABE.
Answer:
(i) Area of parallelogram ABED:
Step 1: Triangle ADE and parallelogram ABED share the same base ED.
Step 2: They lie between the same parallel lines AB and ED.
Step 3: Therefore, Area(//gm ABED) = 2 × Area(ΔADE).
Step 4: Area(//gm ABED) = 2 × 60 = 120 cm².
(ii) Area of rectangle ABCF:
Step 1: Rectangle ABCF and parallelogram ABED share the same base AB.
Step 2: They lie between the same parallel lines AB and FC.
Step 3: Therefore, Area(rect ABCF) = Area(//gm ABED).
Step 4: Area(rect ABCF) = 120 cm².
(iii) Area of triangle ABE:
Step 1: Triangle ABE and parallelogram ABED share the same base AB.
Step 2: They lie between the same parallel lines AB and ED.
Step 3: Therefore, Area(ΔABE) = 1/2 × Area(//gm ABED).
Step 4: Area(ΔABE) = 1/2 × 120 = 60 cm².
(i) Area of parallelogram ABED:
Step 1: Triangle ADE and parallelogram ABED share the same base ED.
Step 2: They lie between the same parallel lines AB and ED.
Step 3: Therefore, Area(//gm ABED) = 2 × Area(ΔADE).
Step 4: Area(//gm ABED) = 2 × 60 = 120 cm².
(ii) Area of rectangle ABCF:
Step 1: Rectangle ABCF and parallelogram ABED share the same base AB.
Step 2: They lie between the same parallel lines AB and FC.
Step 3: Therefore, Area(rect ABCF) = Area(//gm ABED).
Step 4: Area(rect ABCF) = 120 cm².
(iii) Area of triangle ABE:
Step 1: Triangle ABE and parallelogram ABED share the same base AB.
Step 2: They lie between the same parallel lines AB and ED.
Step 3: Therefore, Area(ΔABE) = 1/2 × Area(//gm ABED).
Step 4: Area(ΔABE) = 1/2 × 120 = 60 cm².
3. The given figure shows a rectangle ABDC and a parallelogram ABEF; drawn on opposite sides of AB. Prove that :
(i) quadrilateral CDEF is a parallelogram;
(ii) Area of quad. CDEF = Area of rect. ABDC + Area of //gm. ABEF.
Answer:
(i) Step 1: In rectangle ABDC, side CD is parallel and equal to side AB.
Step 2: In parallelogram ABEF, side FE is parallel and equal to side AB.
Step 3: Therefore, CD is parallel to FE and CD = FE.
Step 4: Since one pair of opposite sides is both parallel and equal, quadrilateral CDEF is a parallelogram.
(ii) Step 1: Let the height of rectangle ABDC be h1 and the height of parallelogram ABEF be h2.
Step 2: The total height of the new parallelogram CDEF is (h1 + h2).
Step 3: Area of parallelogram CDEF = base CD × total height.
Step 4: Area of CDEF = CD × (h1 + h2) = (CD × h1) + (CD × h2).
Step 5: Since CD = AB, we can substitute AB for CD.
Step 6: Area of CDEF = (AB × h1) + (AB × h2).
Step 7: (AB × h1) is the Area of rectangle ABDC and (AB × h2) is the Area of parallelogram ABEF.
Step 8: Therefore, Area of quad. CDEF = Area of rect. ABDC + Area of //gm. ABEF.
(i) Step 1: In rectangle ABDC, side CD is parallel and equal to side AB.
Step 2: In parallelogram ABEF, side FE is parallel and equal to side AB.
Step 3: Therefore, CD is parallel to FE and CD = FE.
Step 4: Since one pair of opposite sides is both parallel and equal, quadrilateral CDEF is a parallelogram.
(ii) Step 1: Let the height of rectangle ABDC be h1 and the height of parallelogram ABEF be h2.
Step 2: The total height of the new parallelogram CDEF is (h1 + h2).
Step 3: Area of parallelogram CDEF = base CD × total height.
Step 4: Area of CDEF = CD × (h1 + h2) = (CD × h1) + (CD × h2).
Step 5: Since CD = AB, we can substitute AB for CD.
Step 6: Area of CDEF = (AB × h1) + (AB × h2).
Step 7: (AB × h1) is the Area of rectangle ABDC and (AB × h2) is the Area of parallelogram ABEF.
Step 8: Therefore, Area of quad. CDEF = Area of rect. ABDC + Area of //gm. ABEF.
4. In the given figure, diagonals PR and QS of the parallelogram PQRS intersect at point O and LM is parallel to PS. Show that :
(i) 2 Area (Δ POS) = Area (//gm PMLS)
(ii) Area (Δ POS) + Area (Δ QOR) = 1/2 Area (//gm PQRS)
(iii) Area (Δ POS) + Area (Δ QOR) = Area (Δ POQ) + Area (Δ SOR).
Answer:
(i) Step 1: Triangle POS and parallelogram PMLS are on the same base PS.
Step 2: They are between the same parallels PS and LM.
Step 3: Therefore, Area(ΔPOS) = 1/2 Area(//gm PMLS).
Step 4: Multiplying by 2, we get 2 Area(ΔPOS) = Area(//gm PMLS).
(ii) Step 1: Similarly, Triangle QOR and parallelogram LMQR are on the same base QR.
Step 2: They are between the same parallels QR and LM.
Step 3: Therefore, 2 Area(ΔQOR) = Area(//gm LMQR).
Step 4: Adding the results from step (i) and step 3:
Step 5: 2 Area(ΔPOS) + 2 Area(ΔQOR) = Area(//gm PMLS) + Area(//gm LMQR).
Step 6: 2 [Area(ΔPOS) + Area(ΔQOR)] = Area(//gm PQRS).
Step 7: Area(ΔPOS) + Area(ΔQOR) = 1/2 Area(//gm PQRS).
(iii) Step 1: The total area of parallelogram PQRS is the sum of the four small triangles.
Step 2: Total Area = Area(ΔPOS) + Area(ΔQOR) + Area(ΔPOQ) + Area(ΔSOR).
Step 3: From part (ii), we know Area(ΔPOS) + Area(ΔQOR) makes up exactly 1/2 of the Total Area.
Step 4: Therefore, the remaining part, Area(ΔPOQ) + Area(ΔSOR), must make up the other 1/2 of the Total Area.
Step 5: Hence, Area(ΔPOS) + Area(ΔQOR) = Area(ΔPOQ) + Area(ΔSOR).
(i) Step 1: Triangle POS and parallelogram PMLS are on the same base PS.
Step 2: They are between the same parallels PS and LM.
Step 3: Therefore, Area(ΔPOS) = 1/2 Area(//gm PMLS).
Step 4: Multiplying by 2, we get 2 Area(ΔPOS) = Area(//gm PMLS).
(ii) Step 1: Similarly, Triangle QOR and parallelogram LMQR are on the same base QR.
Step 2: They are between the same parallels QR and LM.
Step 3: Therefore, 2 Area(ΔQOR) = Area(//gm LMQR).
Step 4: Adding the results from step (i) and step 3:
Step 5: 2 Area(ΔPOS) + 2 Area(ΔQOR) = Area(//gm PMLS) + Area(//gm LMQR).
Step 6: 2 [Area(ΔPOS) + Area(ΔQOR)] = Area(//gm PQRS).
Step 7: Area(ΔPOS) + Area(ΔQOR) = 1/2 Area(//gm PQRS).
(iii) Step 1: The total area of parallelogram PQRS is the sum of the four small triangles.
Step 2: Total Area = Area(ΔPOS) + Area(ΔQOR) + Area(ΔPOQ) + Area(ΔSOR).
Step 3: From part (ii), we know Area(ΔPOS) + Area(ΔQOR) makes up exactly 1/2 of the Total Area.
Step 4: Therefore, the remaining part, Area(ΔPOQ) + Area(ΔSOR), must make up the other 1/2 of the Total Area.
Step 5: Hence, Area(ΔPOS) + Area(ΔQOR) = Area(ΔPOQ) + Area(ΔSOR).
5. In parallelogram ABCD, P is a point on side AB and Q is a point on side BC. Prove that :
(i) Δ CPD and Δ AQD are equal in area.
(ii) Area (Δ AQD) = Area (Δ APD) + Area (Δ CPB).
Answer:
(i) Step 1: Triangle CPD and parallelogram ABCD lie on the same base CD.
Step 2: They lie between the same parallels CD and AB.
Step 3: Therefore, Area(ΔCPD) = 1/2 Area(//gm ABCD).
Step 4: Triangle AQD and parallelogram ABCD lie on the same base AD.
Step 5: They lie between the same parallels AD and BC.
Step 6: Therefore, Area(ΔAQD) = 1/2 Area(//gm ABCD).
Step 7: From steps 3 and 6, Area(ΔCPD) = Area(ΔAQD).
(ii) Step 1: The total Area(//gm ABCD) is comprised of three triangles: ΔCPD, ΔAPD, and ΔCPB.
Step 2: Area(//gm ABCD) = Area(ΔCPD) + Area(ΔAPD) + Area(ΔCPB).
Step 3: We proved Area(ΔCPD) = 1/2 Area(//gm ABCD).
Step 4: Substituting this in, 1/2 Area(//gm ABCD) = Area(ΔAPD) + Area(ΔCPB).
Step 5: We also know Area(ΔAQD) = 1/2 Area(//gm ABCD).
Step 6: Therefore, Area(ΔAQD) = Area(ΔAPD) + Area(ΔCPB).
(i) Step 1: Triangle CPD and parallelogram ABCD lie on the same base CD.
Step 2: They lie between the same parallels CD and AB.
Step 3: Therefore, Area(ΔCPD) = 1/2 Area(//gm ABCD).
Step 4: Triangle AQD and parallelogram ABCD lie on the same base AD.
Step 5: They lie between the same parallels AD and BC.
Step 6: Therefore, Area(ΔAQD) = 1/2 Area(//gm ABCD).
Step 7: From steps 3 and 6, Area(ΔCPD) = Area(ΔAQD).
(ii) Step 1: The total Area(//gm ABCD) is comprised of three triangles: ΔCPD, ΔAPD, and ΔCPB.
Step 2: Area(//gm ABCD) = Area(ΔCPD) + Area(ΔAPD) + Area(ΔCPB).
Step 3: We proved Area(ΔCPD) = 1/2 Area(//gm ABCD).
Step 4: Substituting this in, 1/2 Area(//gm ABCD) = Area(ΔAPD) + Area(ΔCPB).
Step 5: We also know Area(ΔAQD) = 1/2 Area(//gm ABCD).
Step 6: Therefore, Area(ΔAQD) = Area(ΔAPD) + Area(ΔCPB).
6. In the given figure, M and N are the mid-points of the sides DC and AB respectively of the parallelogram ABCD.
If the area of parallelogram ABCD is 48 cm²;
(i) state the area of the triangle BEC.
(ii) name the parallelogram which is equal in area to the triangle BEC.
Answer:
(i) Step 1: Triangle BEC and parallelogram ABCD are on the same base BC.
Step 2: They are between the same parallels BC and AD.
Step 3: Therefore, Area(ΔBEC) = 1/2 Area(//gm ABCD).
Step 4: Area(ΔBEC) = 1/2 × 48 = 24 cm².
(ii) Step 1: Since M and N are midpoints of opposite equal sides DC and AB, AN = DM and AN // DM.
Step 2: Thus, ANMD is a parallelogram.
Step 3: Area(//gm ANMD) = 1/2 Area(//gm ABCD) = 24 cm².
Step 4: Similarly, NBCM is a parallelogram with an area of 24 cm².
Step 5: Therefore, parallelograms ANMD and NBCM are each equal in area to triangle BEC.
(i) Step 1: Triangle BEC and parallelogram ABCD are on the same base BC.
Step 2: They are between the same parallels BC and AD.
Step 3: Therefore, Area(ΔBEC) = 1/2 Area(//gm ABCD).
Step 4: Area(ΔBEC) = 1/2 × 48 = 24 cm².
(ii) Step 1: Since M and N are midpoints of opposite equal sides DC and AB, AN = DM and AN // DM.
Step 2: Thus, ANMD is a parallelogram.
Step 3: Area(//gm ANMD) = 1/2 Area(//gm ABCD) = 24 cm².
Step 4: Similarly, NBCM is a parallelogram with an area of 24 cm².
Step 5: Therefore, parallelograms ANMD and NBCM are each equal in area to triangle BEC.
7. In the following figure, CE is drawn parallel to diagonal DB of the quadrilateral ABCD which meets AB produced at point E. Prove that Δ ADE and quadrilateral ABCD are equal in area.
Answer:
Step 1: Triangle DCB and Triangle DEB are on the same base DB.
Step 2: They are between the same parallels DB and CE.
Step 3: Therefore, Area(ΔDCB) = Area(ΔDEB).
Step 4: Add Area(ΔADB) to both sides of the equation.
Step 5: Area(ΔADB) + Area(ΔDCB) = Area(ΔADB) + Area(ΔDEB).
Step 6: The left side forms the quadrilateral ABCD.
Step 7: The right side forms the triangle ADE.
Step 8: Therefore, Area(quad. ABCD) = Area(ΔADE).
Step 1: Triangle DCB and Triangle DEB are on the same base DB.
Step 2: They are between the same parallels DB and CE.
Step 3: Therefore, Area(ΔDCB) = Area(ΔDEB).
Step 4: Add Area(ΔADB) to both sides of the equation.
Step 5: Area(ΔADB) + Area(ΔDCB) = Area(ΔADB) + Area(ΔDEB).
Step 6: The left side forms the quadrilateral ABCD.
Step 7: The right side forms the triangle ADE.
Step 8: Therefore, Area(quad. ABCD) = Area(ΔADE).
8. ABCD is a parallelogram, a line through A cuts DC at point P and BC produced at Q. Prove that triangle BCP is equal in area to triangle DPQ.
Answer:
Step 1: Triangle ADQ and parallelogram ABCD are on the same base AD and between the same parallels AD and BQ.
Step 2: Therefore, Area(ΔADQ) = 1/2 Area(//gm ABCD).
Step 3: Triangle ABP and parallelogram ABCD are on the same base AB and between the same parallels AB and DC.
Step 4: Therefore, Area(ΔABP) = 1/2 Area(//gm ABCD).
Step 5: Equating the two, Area(ΔADQ) = Area(ΔABP).
Step 6: We can write Area(ΔADQ) as Area(ΔADP) + Area(ΔDPQ).
Step 7: We can write Area(ΔABP) as Area(ΔADP) + Area(ΔDPC) based on subtracting common areas... wait, an easier path is next.
Step 8: Triangle APC and Triangle BPC are on the same base PC and between the same parallels PC and AB.
Step 9: Therefore, Area(ΔAPC) = Area(ΔBPC).
Step 10: Triangle ACQ and Triangle DCQ are on the same base CQ and between the same parallels CQ and AD.
Step 11: So, Area(ΔACQ) = Area(ΔDCQ). Subtracting common Area(ΔPCQ) gives Area(ΔACP) = Area(ΔDPQ).
Step 12: From steps 9 and 11, Area(ΔDPQ) = Area(ΔBCP).
Step 1: Triangle ADQ and parallelogram ABCD are on the same base AD and between the same parallels AD and BQ.
Step 2: Therefore, Area(ΔADQ) = 1/2 Area(//gm ABCD).
Step 3: Triangle ABP and parallelogram ABCD are on the same base AB and between the same parallels AB and DC.
Step 4: Therefore, Area(ΔABP) = 1/2 Area(//gm ABCD).
Step 5: Equating the two, Area(ΔADQ) = Area(ΔABP).
Step 6: We can write Area(ΔADQ) as Area(ΔADP) + Area(ΔDPQ).
Step 7: We can write Area(ΔABP) as Area(ΔADP) + Area(ΔDPC) based on subtracting common areas... wait, an easier path is next.
Step 8: Triangle APC and Triangle BPC are on the same base PC and between the same parallels PC and AB.
Step 9: Therefore, Area(ΔAPC) = Area(ΔBPC).
Step 10: Triangle ACQ and Triangle DCQ are on the same base CQ and between the same parallels CQ and AD.
Step 11: So, Area(ΔACQ) = Area(ΔDCQ). Subtracting common Area(ΔPCQ) gives Area(ΔACP) = Area(ΔDPQ).
Step 12: From steps 9 and 11, Area(ΔDPQ) = Area(ΔBCP).
9. The given figure shows a pentagon ABCDE. EG drawn parallel to DA meets BA produced at G and CF drawn parallel to DB meets AB produced at F. Prove that the area of pentagon ABCDE is equal to the area of triangle GDF.
Answer:
Step 1: Triangle EDA and Triangle GDA are on the same base DA.
Step 2: They are between the same parallels DA and EG.
Step 3: Therefore, Area(ΔEDA) = Area(ΔGDA).
Step 4: Triangle CDB and Triangle FDB are on the same base DB.
Step 5: They are between the same parallels DB and CF.
Step 6: Therefore, Area(ΔCDB) = Area(ΔFDB).
Step 7: Area of pentagon ABCDE = Area(ΔEDA) + Area(ΔADB) + Area(ΔCDB).
Step 8: Substitute the equal areas from step 3 and 6.
Step 9: Area of pentagon ABCDE = Area(ΔGDA) + Area(ΔADB) + Area(ΔFDB).
Step 10: The sum of these three areas is exactly the Area of ΔGDF.
Step 11: Therefore, Area of pentagon ABCDE = Area(ΔGDF).
Step 1: Triangle EDA and Triangle GDA are on the same base DA.
Step 2: They are between the same parallels DA and EG.
Step 3: Therefore, Area(ΔEDA) = Area(ΔGDA).
Step 4: Triangle CDB and Triangle FDB are on the same base DB.
Step 5: They are between the same parallels DB and CF.
Step 6: Therefore, Area(ΔCDB) = Area(ΔFDB).
Step 7: Area of pentagon ABCDE = Area(ΔEDA) + Area(ΔADB) + Area(ΔCDB).
Step 8: Substitute the equal areas from step 3 and 6.
Step 9: Area of pentagon ABCDE = Area(ΔGDA) + Area(ΔADB) + Area(ΔFDB).
Step 10: The sum of these three areas is exactly the Area of ΔGDF.
Step 11: Therefore, Area of pentagon ABCDE = Area(ΔGDF).
10. In the given figure, AP is parallel to BC, BP is parallel to CQ. Prove that the areas of triangles ABC and BQP are equal.
Answer:
Step 1: Join points P and C.
Step 2: Triangle ABC and Triangle PBC are on the same base BC.
Step 3: They are between the same parallel lines BC and AP.
Step 4: Therefore, Area(ΔABC) = Area(ΔPBC).
Step 5: Triangle PBC and Triangle PBQ are on the same base BP.
Step 6: They are between the same parallel lines BP and CQ.
Step 7: Therefore, Area(ΔPBC) = Area(ΔPBQ).
Step 8: Combining results from step 4 and step 7, Area(ΔABC) = Area(ΔBQP).
Step 1: Join points P and C.
Step 2: Triangle ABC and Triangle PBC are on the same base BC.
Step 3: They are between the same parallel lines BC and AP.
Step 4: Therefore, Area(ΔABC) = Area(ΔPBC).
Step 5: Triangle PBC and Triangle PBQ are on the same base BP.
Step 6: They are between the same parallel lines BP and CQ.
Step 7: Therefore, Area(ΔPBC) = Area(ΔPBQ).
Step 8: Combining results from step 4 and step 7, Area(ΔABC) = Area(ΔBQP).
11. In the figure given alongside, squares ABDE and AFGC are drawn on the side AB and the hypotenuse AC of the right triangle ABC. If BH is perpendicular to FG, prove that :
(i) Δ EAC ≅ Δ BAF.
(ii) Area of the square ABDE = Area of the rectangle ARHF.
Answer:
(i) Step 1: In ΔEAC and ΔBAF, side EA = side BA (sides of square ABDE).
Step 2: Side AC = side AF (sides of square AFGC).
Step 3: Angle EAC = 90° + Angle BAC.
Step 4: Angle BAF = 90° + Angle BAC.
Step 5: Therefore, Angle EAC = Angle BAF.
Step 6: By SAS congruence criterion, ΔEAC ≅ ΔBAF.
(ii) Step 1: Triangle EAC and square ABDE are on the same base EA and between same parallels EA and DC (extending DB).
Step 2: Therefore, Area(ΔEAC) = 1/2 Area(square ABDE).
Step 3: Triangle BAF and rectangle ARHF are on the same base AF and between same parallels AF and BH.
Step 4: Therefore, Area(ΔBAF) = 1/2 Area(rectangle ARHF).
Step 5: Since ΔEAC ≅ ΔBAF, their areas are equal.
Step 6: Therefore, 1/2 Area(square ABDE) = 1/2 Area(rectangle ARHF).
Step 7: Thus, Area(square ABDE) = Area(rectangle ARHF).
(i) Step 1: In ΔEAC and ΔBAF, side EA = side BA (sides of square ABDE).
Step 2: Side AC = side AF (sides of square AFGC).
Step 3: Angle EAC = 90° + Angle BAC.
Step 4: Angle BAF = 90° + Angle BAC.
Step 5: Therefore, Angle EAC = Angle BAF.
Step 6: By SAS congruence criterion, ΔEAC ≅ ΔBAF.
(ii) Step 1: Triangle EAC and square ABDE are on the same base EA and between same parallels EA and DC (extending DB).
Step 2: Therefore, Area(ΔEAC) = 1/2 Area(square ABDE).
Step 3: Triangle BAF and rectangle ARHF are on the same base AF and between same parallels AF and BH.
Step 4: Therefore, Area(ΔBAF) = 1/2 Area(rectangle ARHF).
Step 5: Since ΔEAC ≅ ΔBAF, their areas are equal.
Step 6: Therefore, 1/2 Area(square ABDE) = 1/2 Area(rectangle ARHF).
Step 7: Thus, Area(square ABDE) = Area(rectangle ARHF).
12. In the following figure, DE is parallel to BC. Show that :
(i) Area (Δ ADC) = Area (Δ AEB)
(ii) Area (Δ BOD) = Area (Δ COE).
Answer:
(i) Step 1: Triangle DBC and Triangle EBC are on the same base BC.
Step 2: They lie between the same parallel lines BC and DE.
Step 3: Therefore, Area(ΔDBC) = Area(ΔEBC).
Step 4: Subtract these areas from the total Area(ΔABC).
Step 5: Area(ΔABC) - Area(ΔDBC) = Area(ΔABC) - Area(ΔEBC).
Step 6: Area(ΔADC) = Area(ΔAEB).
(ii) Step 1: We established that Area(ΔDBC) = Area(ΔEBC).
Step 2: Both these triangles share the common triangle OBC.
Step 3: Subtract Area(ΔOBC) from both sides.
Step 4: Area(ΔDBC) - Area(ΔOBC) = Area(ΔEBC) - Area(ΔOBC).
Step 5: Area(ΔBOD) = Area(ΔCOE).
(i) Step 1: Triangle DBC and Triangle EBC are on the same base BC.
Step 2: They lie between the same parallel lines BC and DE.
Step 3: Therefore, Area(ΔDBC) = Area(ΔEBC).
Step 4: Subtract these areas from the total Area(ΔABC).
Step 5: Area(ΔABC) - Area(ΔDBC) = Area(ΔABC) - Area(ΔEBC).
Step 6: Area(ΔADC) = Area(ΔAEB).
(ii) Step 1: We established that Area(ΔDBC) = Area(ΔEBC).
Step 2: Both these triangles share the common triangle OBC.
Step 3: Subtract Area(ΔOBC) from both sides.
Step 4: Area(ΔDBC) - Area(ΔOBC) = Area(ΔEBC) - Area(ΔOBC).
Step 5: Area(ΔBOD) = Area(ΔCOE).
13. Show that :
(i) a diagonal divides a parallelogram into two triangles of equal area.
(ii) the ratio of the areas of two triangles of the same height is equal to the ratio of their bases.
(iii) the ratio of the areas of two triangles on the same base is equal to the ratio of their heights.
Answer:
(i) Step 1: Let ABCD be a parallelogram and AC be its diagonal.
Step 2: In ΔABC and ΔCDA, AB = CD and BC = DA (opposite sides of a parallelogram).
Step 3: AC is common to both.
Step 4: By SSS congruence criterion, ΔABC ≅ ΔCDA.
Step 5: Since congruent triangles have equal areas, Area(ΔABC) = Area(ΔCDA).
(ii) Step 1: Let two triangles have bases B1 and B2, and the same height H.
Step 2: Area 1 = 1/2 × B1 × H.
Step 3: Area 2 = 1/2 × B2 × H.
Step 4: Ratio = (1/2 × B1 × H) / (1/2 × B2 × H) = B1 / B2.
Step 5: This proves the ratio of areas is equal to the ratio of their bases.
(iii) Step 1: Let two triangles have the same base B, and heights H1 and H2.
Step 2: Area 1 = 1/2 × B × H1.
Step 3: Area 2 = 1/2 × B × H2.
Step 4: Ratio = (1/2 × B × H1) / (1/2 × B × H2) = H1 / H2.
Step 5: This proves the ratio of areas is equal to the ratio of their heights.
(i) Step 1: Let ABCD be a parallelogram and AC be its diagonal.
Step 2: In ΔABC and ΔCDA, AB = CD and BC = DA (opposite sides of a parallelogram).
Step 3: AC is common to both.
Step 4: By SSS congruence criterion, ΔABC ≅ ΔCDA.
Step 5: Since congruent triangles have equal areas, Area(ΔABC) = Area(ΔCDA).
(ii) Step 1: Let two triangles have bases B1 and B2, and the same height H.
Step 2: Area 1 = 1/2 × B1 × H.
Step 3: Area 2 = 1/2 × B2 × H.
Step 4: Ratio = (1/2 × B1 × H) / (1/2 × B2 × H) = B1 / B2.
Step 5: This proves the ratio of areas is equal to the ratio of their bases.
(iii) Step 1: Let two triangles have the same base B, and heights H1 and H2.
Step 2: Area 1 = 1/2 × B × H1.
Step 3: Area 2 = 1/2 × B × H2.
Step 4: Ratio = (1/2 × B × H1) / (1/2 × B × H2) = H1 / H2.
Step 5: This proves the ratio of areas is equal to the ratio of their heights.
14. In the given figure; AD is median of Δ ABC and E is any point on median AD. Prove that Area (Δ ABE) = Area (Δ ACE).
Answer:
Step 1: AD is the median of ΔABC, so it divides it into equal areas.
Step 2: Area(ΔABD) = Area(ΔACD).
Step 3: D is the midpoint of BC, so ED is the median of ΔEBC.
Step 4: Therefore, Area(ΔEBD) = Area(ΔECD).
Step 5: Subtract the equation in step 4 from the equation in step 2.
Step 6: Area(ΔABD) - Area(ΔEBD) = Area(ΔACD) - Area(ΔECD).
Step 7: Area(ΔABE) = Area(ΔACE).
Step 1: AD is the median of ΔABC, so it divides it into equal areas.
Step 2: Area(ΔABD) = Area(ΔACD).
Step 3: D is the midpoint of BC, so ED is the median of ΔEBC.
Step 4: Therefore, Area(ΔEBD) = Area(ΔECD).
Step 5: Subtract the equation in step 4 from the equation in step 2.
Step 6: Area(ΔABD) - Area(ΔEBD) = Area(ΔACD) - Area(ΔECD).
Step 7: Area(ΔABE) = Area(ΔACE).
15. In the figure of question 14, if E is the mid point of median AD, then prove that : Area (Δ ABE) = 1/4 Area (Δ ABC).
Answer:
Step 1: Since AD is the median of ΔABC, Area(ΔABD) = 1/2 Area(ΔABC).
Step 2: In ΔABD, BE is the median because E is the midpoint of AD.
Step 3: A median divides a triangle into two equal areas, so Area(ΔABE) = 1/2 Area(ΔABD).
Step 4: Substitute the value from step 1 into this equation.
Step 5: Area(ΔABE) = 1/2 × [1/2 Area(ΔABC)].
Step 6: Area(ΔABE) = 1/4 Area(ΔABC).
Step 1: Since AD is the median of ΔABC, Area(ΔABD) = 1/2 Area(ΔABC).
Step 2: In ΔABD, BE is the median because E is the midpoint of AD.
Step 3: A median divides a triangle into two equal areas, so Area(ΔABE) = 1/2 Area(ΔABD).
Step 4: Substitute the value from step 1 into this equation.
Step 5: Area(ΔABE) = 1/2 × [1/2 Area(ΔABC)].
Step 6: Area(ΔABE) = 1/4 Area(ΔABC).
16. ABCD is a parallelogram. P and Q are the mid-points of sides AB and AD respectively. Prove that area of triangle APQ = 1/8 of the area of parallelogram ABCD.
Answer:
Step 1: Join points B and D, and points P and D.
Step 2: In ΔABD, P is the midpoint of AB, making DP a median.
Step 3: Therefore, Area(ΔAPD) = 1/2 Area(ΔABD).
Step 4: In ΔAPD, Q is the midpoint of AD, making PQ a median.
Step 5: Therefore, Area(ΔAPQ) = 1/2 Area(ΔAPD) = 1/4 Area(ΔABD).
Step 6: The diagonal BD divides parallelogram ABCD into two equal triangles, so Area(ΔABD) = 1/2 Area(//gm ABCD).
Step 7: Substitute this into the previous step.
Step 8: Area(ΔAPQ) = 1/4 × [1/2 Area(//gm ABCD)] = 1/8 Area(//gm ABCD).
Step 1: Join points B and D, and points P and D.
Step 2: In ΔABD, P is the midpoint of AB, making DP a median.
Step 3: Therefore, Area(ΔAPD) = 1/2 Area(ΔABD).
Step 4: In ΔAPD, Q is the midpoint of AD, making PQ a median.
Step 5: Therefore, Area(ΔAPQ) = 1/2 Area(ΔAPD) = 1/4 Area(ΔABD).
Step 6: The diagonal BD divides parallelogram ABCD into two equal triangles, so Area(ΔABD) = 1/2 Area(//gm ABCD).
Step 7: Substitute this into the previous step.
Step 8: Area(ΔAPQ) = 1/4 × [1/2 Area(//gm ABCD)] = 1/8 Area(//gm ABCD).
17. The base BC of triangle ABC is divided at D so that BD = 1/2 DC. Prove that area of Δ ABD = 1/3 of the area of Δ ABC.
Answer:
Step 1: Given BD = 1/2 DC, this can be written as DC = 2 BD.
Step 2: The entire base BC = BD + DC = BD + 2 BD = 3 BD.
Step 3: Therefore, the ratio of bases BD / BC = 1 / 3.
Step 4: Triangles ABD and ABC share the same vertex A, so their heights are equal.
Step 5: The ratio of their areas equals the ratio of their bases.
Step 6: Area(ΔABD) / Area(ΔABC) = BD / BC = 1 / 3.
Step 7: Area(ΔABD) = 1/3 Area(ΔABC).
Step 1: Given BD = 1/2 DC, this can be written as DC = 2 BD.
Step 2: The entire base BC = BD + DC = BD + 2 BD = 3 BD.
Step 3: Therefore, the ratio of bases BD / BC = 1 / 3.
Step 4: Triangles ABD and ABC share the same vertex A, so their heights are equal.
Step 5: The ratio of their areas equals the ratio of their bases.
Step 6: Area(ΔABD) / Area(ΔABC) = BD / BC = 1 / 3.
Step 7: Area(ΔABD) = 1/3 Area(ΔABC).
18. In a parallelogram ABCD, point P lies in DC such that DP : PC = 3 : 2. If area of Δ DPB = 30 sq. cm, find the area of the parallelogram ABCD.
Answer:
Step 1: Triangles DPB and PCB share vertex B, with bases on the same line DC.
Step 2: The ratio of their areas equals the ratio of their bases (DP : PC = 3 : 2).
Step 3: Area(ΔDPB) / Area(ΔPCB) = 3 / 2.
Step 4: 30 / Area(ΔPCB) = 3 / 2, which gives Area(ΔPCB) = 20 sq. cm.
Step 5: The total area of ΔBDC = Area(ΔDPB) + Area(ΔPCB) = 30 + 20 = 50 sq. cm.
Step 6: The diagonal BD divides the parallelogram into two equal halves.
Step 7: Area(//gm ABCD) = 2 × Area(ΔBDC) = 2 × 50 = 100 sq. cm.
Step 1: Triangles DPB and PCB share vertex B, with bases on the same line DC.
Step 2: The ratio of their areas equals the ratio of their bases (DP : PC = 3 : 2).
Step 3: Area(ΔDPB) / Area(ΔPCB) = 3 / 2.
Step 4: 30 / Area(ΔPCB) = 3 / 2, which gives Area(ΔPCB) = 20 sq. cm.
Step 5: The total area of ΔBDC = Area(ΔDPB) + Area(ΔPCB) = 30 + 20 = 50 sq. cm.
Step 6: The diagonal BD divides the parallelogram into two equal halves.
Step 7: Area(//gm ABCD) = 2 × Area(ΔBDC) = 2 × 50 = 100 sq. cm.
TEST YOURSELF
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) The median of a triangle divides it into two:
(i) triangles of equal area
(ii) congruent triangles
(iii) right triangles
(iv) isosceles triangles
Answer: (i) triangles of equal area
Step 1: The median bisects the base.
Step 2: Both triangles share the same height.
Step 3: Therefore, their areas are equal.
Step 1: The median bisects the base.
Step 2: Both triangles share the same height.
Step 3: Therefore, their areas are equal.
(b) The area of given parallelogram is :
(i) AB × BM
(ii) BC × BN
(iii) DC × DL
(iv) AD × DL
Answer: (iii) DC × DL
Step 1: The area of a parallelogram is Base × Altitude.
Step 2: Looking at the figure, DL is the perpendicular altitude corresponding to the base AB (or DC, since AB = DC).
Step 3: Area = Base × Height = DC × DL.
Step 1: The area of a parallelogram is Base × Altitude.
Step 2: Looking at the figure, DL is the perpendicular altitude corresponding to the base AB (or DC, since AB = DC).
Step 3: Area = Base × Height = DC × DL.
(c) ABCD is a quadrilateral whose diagonals intersect each other at point O. The diagonal AC bisects diagonal BD. Then area of quadrilateral ABCD is :
(i) 2 × area of ΔABD
(ii) 2 × area of ΔBCD
(iii) 4 × area of ΔAOB
(iv) 2 × area of ΔABC
Answer: (iv) 2 × area of ΔABC
Step 1: Since AC bisects BD, O is the midpoint of BD.
Step 2: AO is the median for ΔABD, so Area(ΔAOB) = Area(ΔAOD).
Step 3: CO is the median for ΔBCD, so Area(ΔBOC) = Area(ΔCOD).
Step 4: Total Area = Area(ΔAOB) + Area(ΔAOD) + Area(ΔBOC) + Area(ΔCOD).
Step 5: Total Area = 2 × Area(ΔAOB) + 2 × Area(ΔBOC) = 2 × [Area(ΔAOB) + Area(ΔBOC)].
Step 6: Total Area = 2 × Area(ΔABC).
Step 1: Since AC bisects BD, O is the midpoint of BD.
Step 2: AO is the median for ΔABD, so Area(ΔAOB) = Area(ΔAOD).
Step 3: CO is the median for ΔBCD, so Area(ΔBOC) = Area(ΔCOD).
Step 4: Total Area = Area(ΔAOB) + Area(ΔAOD) + Area(ΔBOC) + Area(ΔCOD).
Step 5: Total Area = 2 × Area(ΔAOB) + 2 × Area(ΔBOC) = 2 × [Area(ΔAOB) + Area(ΔBOC)].
Step 6: Total Area = 2 × Area(ΔABC).
(d) Two parallelograms ABCD and ABEF are equal in area, they lie between the same parallel lines :
(i) Yes
(ii) No
(iii) Nothing can be said
Answer: (iii) Nothing can be said
Step 1: Two parallelograms on the same base with equal areas must have equal heights.
Step 2: However, they could lie on opposite sides of the shared base AB.
Step 3: Unless we know they are on the same side of the base, we cannot guarantee they are between the exact same parallel lines.
Step 1: Two parallelograms on the same base with equal areas must have equal heights.
Step 2: However, they could lie on opposite sides of the shared base AB.
Step 3: Unless we know they are on the same side of the base, we cannot guarantee they are between the exact same parallel lines.
(e) ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. E and F are mid-points of the non-parallel sides. The ratio of ar. (ABFE) and ar.(EFCD) is :
(i) a : b
(ii) (3a + b) : (a + 3b)
(iii) (a + 3b) : (3a + b)
(iv) (2a + b) : (3a + b)
Answer: (ii) (3a + b) : (a + 3b)
Step 1: The length of the midsegment EF = (a + b) / 2.
Step 2: Let the total height of the trapezium be 2h. The height of each smaller trapezium is h.
Step 3: Area(ABFE) = 1/2 × [a + (a + b)/2] × h = 1/2 × [(3a + b)/2] × h.
Step 4: Area(EFCD) = 1/2 × [b + (a + b)/2] × h = 1/2 × [(a + 3b)/2] × h.
Step 5: The ratio is (3a + b) : (a + 3b).
Step 1: The length of the midsegment EF = (a + b) / 2.
Step 2: Let the total height of the trapezium be 2h. The height of each smaller trapezium is h.
Step 3: Area(ABFE) = 1/2 × [a + (a + b)/2] × h = 1/2 × [(3a + b)/2] × h.
Step 4: Area(EFCD) = 1/2 × [b + (a + b)/2] × h = 1/2 × [(a + 3b)/2] × h.
Step 5: The ratio is (3a + b) : (a + 3b).
(f) Statement (1) : ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area. Statement (2) : It is not necessary that the quadrilateral ABCD is a rectangle or a parallelogram or rhombus.
(i) Both the statements are true.
(ii) Both the statements are false.
(iii) Statement 1 is true, and statement 2 is false.
(iv) Statement 1 is false, and statement 2 is true.
Answer: (i) Both the statements are true.
Step 1: A kite is an example of a quadrilateral where a diagonal divides it into two equal triangles.
Step 2: A kite is not a rectangle, parallelogram, or rhombus.
Step 3: Therefore, both statements are factually correct.
Step 1: A kite is an example of a quadrilateral where a diagonal divides it into two equal triangles.
Step 2: A kite is not a rectangle, parallelogram, or rhombus.
Step 3: Therefore, both statements are factually correct.
(g) Assertion (A) : PQRS a parallelogram whose area is 180 cm² and A is any point on the diagonal PR. The area of triangle ASR = 45 cm². Reason (R) : A is not the mid-point of diagonal PR.
(i) A is true, R is false.
(ii) A is false, R is true.
(iii) Both A and R are true and R is the correct reason for A.
(iv) Both A and R are true and R is the incorrect reason for A.
Answer: (ii) A is false, R is true.
Step 1: The area of triangle PSR is half of 180 = 90 cm².
Step 2: For Area(ΔASR) to be 45 cm², A would HAVE to be the exact midpoint of PR.
Step 3: The assertion says A is "any point", which means the area is not always 45 cm², making A false.
Step 4: Because A is an arbitrary point, it is not necessarily the midpoint, making R true.
Step 1: The area of triangle PSR is half of 180 = 90 cm².
Step 2: For Area(ΔASR) to be 45 cm², A would HAVE to be the exact midpoint of PR.
Step 3: The assertion says A is "any point", which means the area is not always 45 cm², making A false.
Step 4: Because A is an arbitrary point, it is not necessarily the midpoint, making R true.
(h) Assertion (A) : ABCD is a square. E is mid-point of side AB and F is mid-point of side DC. If DA = 16 cm, the area of triangle COF is 32 cm². Reason (R) : EF is ⊥ to DC and OF = 1/2 EF = 1/2 DA = 8 cm. Area of COF = 1/2 × CF × OF.
(i) A is true, R is false.
(ii) A is false, R is true.
(iii) Both A and R are true and R is the correct reason for A.
(iv) Both A and R are true and R is the incorrect reason for A.
Answer: (iii) Both A and R are true and R is the correct reason for A.
Step 1: DC = 16. Because F is the midpoint, CF = 8 cm.
Step 2: The line EF joins the midpoints, so EF = 16 cm. Since diagonals of a square intersect at center O, OF = 8 cm.
Step 3: Area of ΔCOF = 1/2 × Base × Height = 1/2 × 8 × 8 = 32 cm².
Step 4: This matches the Assertion perfectly, and the Reason precisely details this calculation.
Step 1: DC = 16. Because F is the midpoint, CF = 8 cm.
Step 2: The line EF joins the midpoints, so EF = 16 cm. Since diagonals of a square intersect at center O, OF = 8 cm.
Step 3: Area of ΔCOF = 1/2 × Base × Height = 1/2 × 8 × 8 = 32 cm².
Step 4: This matches the Assertion perfectly, and the Reason precisely details this calculation.
2. ABCD and BCFE are parallelograms. If area of triangle EBC = 480 cm², AB = 30 cm and BC = 40 cm; Calculate;
(i) area of parallelogram ABCD;
(ii) area of the parallelogram BCFE;
(iii) length of altitude from A on CD;
(iv) area of triangle ECF.
Answer:
(ii) Area of parallelogram BCFE:
Step 1: Triangle EBC and parallelogram BCFE are on the same base BC and between same parallels.
Step 2: Area(//gm BCFE) = 2 × Area(ΔEBC) = 2 × 480 = 960 cm².
(i) Area of parallelogram ABCD:
Step 1: Parallelograms ABCD and BCFE share the same base BC and lie between same parallels.
Step 2: Area(//gm ABCD) = Area(//gm BCFE) = 960 cm².
(iii) Length of altitude from A on CD:
Step 1: Area of //gm ABCD = Base CD × Altitude.
Step 2: Since it's a parallelogram, CD = AB = 30 cm.
Step 3: 960 = 30 × Altitude.
Step 4: Altitude = 960 / 30 = 32 cm.
(iv) Area of triangle ECF:
Step 1: The diagonal EC divides parallelogram BCFE into two equal triangles, ΔEBC and ΔECF.
Step 2: Area(ΔECF) = 1/2 Area(//gm BCFE) = 480 cm².
(ii) Area of parallelogram BCFE:
Step 1: Triangle EBC and parallelogram BCFE are on the same base BC and between same parallels.
Step 2: Area(//gm BCFE) = 2 × Area(ΔEBC) = 2 × 480 = 960 cm².
(i) Area of parallelogram ABCD:
Step 1: Parallelograms ABCD and BCFE share the same base BC and lie between same parallels.
Step 2: Area(//gm ABCD) = Area(//gm BCFE) = 960 cm².
(iii) Length of altitude from A on CD:
Step 1: Area of //gm ABCD = Base CD × Altitude.
Step 2: Since it's a parallelogram, CD = AB = 30 cm.
Step 3: 960 = 30 × Altitude.
Step 4: Altitude = 960 / 30 = 32 cm.
(iv) Area of triangle ECF:
Step 1: The diagonal EC divides parallelogram BCFE into two equal triangles, ΔEBC and ΔECF.
Step 2: Area(ΔECF) = 1/2 Area(//gm BCFE) = 480 cm².
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Quick Review Flashcards - Click to flip and test your knowledge!
Question
How is the area of a plane figure defined?
Answer
The region bounded by the figure.
Question
What is the relationship between the areas of two congruent figures?
Answer
They are always equal in area.
Question
Is it always true that figures with equal areas are congruent?
Answer
No, the converse of the congruency-area rule is not always true.
Question
When are figures said to be 'between the same parallels'?
Answer
When their bases lie on the same straight line and their opposite vertices lie on a parallel straight line.
Question
Figures between the same parallels always have equal _____.
Answer
Altitudes.
Question
What is the formula for the area of a parallelogram?
Answer
$Base \times height$
Question
What is the formula for the area of a triangle?
Answer
$\frac{1}{2} \times base \times height$
Question
According to Theorem 19, parallelograms on the same base and between the same parallels are _____.
Answer
Equal in area.
Question
What is the relationship between a parallelogram and a rectangle on the same base and between the same parallels?
Answer
They are equal in area.
Question
Theorem 20: The area of a triangle is _____ the area of a parallelogram on the same base and between the same parallels.
Answer
Half
Question
Theorem 21: Triangles on the same base and between the same parallels are _____.
Answer
Equal in area.
Question
If two parallelograms stand on equal bases and lie between the same parallels, how do their areas compare?
Answer
They are equal in area.
Question
How does the area of a triangle compare to a parallelogram on an equal base and between the same parallels?
Answer
The area of the triangle is half that of the parallelogram.
Question
Under what condition regarding base and area will two triangles have equal corresponding altitudes?
Answer
They must have equal areas and stand on the same base (or equal bases).
Question
In the proof of Theorem 19, what criterion is used to prove $\triangle ADF \cong \triangle BCE$?
Answer
$A.S.A.$ (Angle-Side-Angle).
Question
What effect does a diagonal have on the area of a parallelogram?
Answer
It bisects the parallelogram into two triangles of equal area.
Question
If a point $P$ lies inside a parallelogram $ABCD$, what is $Area(\triangle APB) + Area(\triangle CPD)$ equal to?
Answer
$\frac{1}{2} Area(Parallelogram \ ABCD)$
Question
Into what does a median divide a triangle?
Answer
Two triangles of equal area.
Question
If $AD$ divides base $BC$ of $\triangle ABC$ in the ratio $m:n$, what is the ratio of $Area(\triangle ABD)$ to $Area(\triangle ADC)$?
Answer
$m:n$
Question
Triangles with the same vertex and bases along the same line have areas in the ratio of their _____.
Answer
Bases.
Question
In $\triangle ABC$, if point $D$ on side $BC$ results in $2BD = 3DC$, what is the ratio of $Area(\triangle ABD)$ to $Area(\triangle ABC)$?
Answer
$\frac{3}{5}$
Question
In parallelogram $ABCD$, if points $P$ and $Q$ trisect side $BC$, what is $Area(\triangle APQ)$ relative to the parallelogram?
Answer
$\frac{1}{6} Area(Parallelogram \ ABCD)$
Question
For a quadrilateral $ABCD$ with diagonals intersecting at $O$, what is the product $Area(\triangle AOD) \times Area(\triangle BOC)$ equal to?
Answer
$Area(\triangle AOB) \times Area(\triangle COD)$
Question
If $BD : DC = 3 : 5$ in $\triangle ABC$, what is the ratio $Area(\triangle ABD) : Area(\triangle ACD)$?
Answer
$3 : 5$
Question
What is the relationship between $Area(\triangle AOD)$ and $Area(\triangle BOC)$ in a trapezium where $AB \parallel DC$ and diagonals intersect at $O$?
Answer
They are equal in area.
Question
If $E$ is the mid-point of side $BC$ in $\triangle ABC$, how does $Area(\triangle BEA)$ compare to $Area(\triangle CEA)$?
Answer
They are equal in area.
Question
In $\triangle ABC$, if $AD$ is the median and $E$ is the midpoint of $AD$, then $Area(\triangle ABE) = \dots Area(\triangle ABC)$.
Answer
$\frac{1}{4}$
Question
If $ABCD$ is a parallelogram and $P$ and $Q$ are midpoints of $AB$ and $AD$, what is the area of $\triangle APQ$ relative to the parallelogram?
Answer
$\frac{1}{8}$ of the area of the parallelogram.
Question
If $BC$ of $\triangle ABC$ is divided at $D$ such that $BD = \frac{1}{2}DC$, what fraction of $Area(\triangle ABC)$ is $Area(\triangle ABD)$?
Answer
$\frac{1}{3}$
Question
In parallelogram $ABCD$, if $DP : PC = 3 : 2$, what is the ratio of $Area(\triangle DPB)$ to $Area(\triangle BPC)$?
Answer
$3 : 2$
Question
In a trapezium with parallel sides $a$ and $b$, the median line joining midpoints of non-parallel sides divides it into two trapeziums with area ratio _____.
Answer
$(3a + b) : (a + 3b)$
Question
In parallelogram $ABCD$, if $E$ is a point on $AB$ such that $AE : EB = 4 : 5$, what is the ratio of $Area(\triangle ADE)$ to $Area(\triangle ADB)$?
Answer
$4 : 9$
Question
What is the relationship between the areas of $\triangle ADE$ and quadrilateral $ABCD$ if $CE$ is drawn parallel to diagonal $DB$ of quadrilateral $ABCD$?
Answer
They are equal in area.
Question
Assertion: $PQRS$ is a parallelogram with area $180 \text{ cm}^2$. If $A$ is any point on diagonal $PR$, what is the area of $\triangle ASR$?
Answer
$45 \text{ cm}^2$
Question
If the perimeter of $\triangle ABC$ is $37 \text{ cm}$ and altitudes are in ratio $6 : 5 : 4$, the ratio of the lengths of its sides is _____.
Answer
$\frac{1}{6} : \frac{1}{5} : \frac{1}{4}$ (or $10 : 12 : 15$)
Question
The ratio of the areas of two triangles on the same base is equal to the ratio of their _____.
Answer
Heights.
Question
The ratio of the areas of two triangles of the same height is equal to the ratio of their _____.
Answer
Bases.
Question
In parallelogram $ABCD$, if $P$ and $Q$ are midpoints of $BC$ and $CD$, then $Area(\triangle APQ)$ is _____ of $Area(Parallelogram \ ABCD)$.
Answer
$\frac{3}{8}$
Question
A line through the vertex $A$ of $\triangle ABC$ meets $BC$ at $D$ and the median through $B$ at $E$. If $AE : ED = 2 : 1$, then $D$ is the _____ of $BC$.
Answer
Midpoint.
Question
If the medians of $\triangle ABC$ intersect at $G$, then $Area(\triangle BGC)$ is equal to _____ $Area(\triangle ABC)$.
Answer
$\frac{1}{3}$
Question
In $\triangle ABC$, if $G$ is the centroid, then $Area(\triangle AGB) = Area(\triangle BGC) = \dots$
Answer
$Area(\triangle AGC)$
Question
If $ABCD$ is a parallelogram, any line through the intersection of its diagonals divides the parallelogram into two parts of _____ area.
Answer
Equal
Question
In a parallelogram $ABCD$, if $P$ is the midpoint of $DC$, how does $Area(\triangle APB)$ relate to the area of the parallelogram?
Answer
It is half the area of the parallelogram.
Question
Theorem: If two triangles have equal base and equal area, they lie between the _____ parallels.
Answer
Same
Question
In $\triangle ABC$, if $D$ and $E$ are points on $AB$ and $AC$ such that $DE \parallel BC$, then $Area(\triangle BCD)$ is _____ $Area(\triangle BCE)$.
Answer
Equal to
Question
How does the area of a rectangle compare to a parallelogram if they have the same base and the same height?
Answer
They are equal in area.
Question
In quadrilateral $ABCD$, if $AC$ bisects $BD$, then $Area(\triangle ABC) = \dots$
Answer
$Area(\triangle ADC)$
Question
If the ratio of altitudes of a triangle is $a:b:c$, the ratio of the corresponding sides is _____.
Answer
$\frac{1}{a} : \frac{1}{b} : \frac{1}{c}$
Question
A median of a triangle divides it into two triangles of equal area because they have the same _____ and equal _____.
Answer
Altitude; bases.
Question
What is the area of a square with side $DA = 16 \text{ cm}$?
Answer
$256 \text{ cm}^2$
Question
If $E, F, G, H$ are mid-points of the sides of parallelogram $ABCD$, then $Area(EFGH)$ is _____ $Area(ABCD)$.
Answer
Half