Chapter 15 Areas of Parallelogram and Triangles R.D. Sharma Solutions for Class 9th Exercise 15.3
Exercise 15.3
1. In Fig. 15.74, compute the area of quadrilateral ABCD.
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Solution
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2. In Fig. 15.75, PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of Δ OTS if PQ = 8 cm.
Solution
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3. Compute the area of trapezium PQRS in figure
Solution
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4. In Fig. 15.77, ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Δ AOB.
Solution
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5. In Fig. 15.78, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.
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Solution
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6. In Fig. 15.79, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 2√5, find the area of the rectangle.
Solution
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7. In Fig. 15.80, ABCD is a trapezium in which AB || DC. Prove that:
ar(Δ AOD) = ar(Δ BOC)
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Solution
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8. In Fig. 15.81, ABCD, ABFE and CDEF are parallelograms. Prove that
ar(Δ ADE) = ar(Δ BCF)
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Solution
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9. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(Δ APB) ✕ ar (Δ CPD) = ar (Δ APD) ✕ ar (Δ BPC)
Solution
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10. In Fig. 15.82, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar (Δ ABC) = ar (Δ ABD)
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11. If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.
Solution
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12. If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of median AD, prove that ar(Δ BGC)= 2 ar (Δ AGC).
Solution
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13. A point D is taken on the side BC of a ΔABC such that BD = 2DC. Prove that ar (Δ ABD) = 2 ar (Δ ADC).
Solution
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14. ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that:
(i) ar (Δ ADO) = ar(Δ CDO)
(ii) ar (Δ ABP) = ar (Δ CBP).
Solution
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Solution
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16. ABCD is a parallelogram whose diagonals AC and BD intersect at O . A line through O intersects AB at P and DC at Q. Prove that ar(ΔPOA)=ar(ΔQOC).
Solution
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17. ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.
Solution
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Solution
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20. In Fig. 15.83, CD || AE and CY || BA.
(i) Name a triangle equal in area of ΔCBX
(ii) Prove that ar (Δ ZDE) = ar (Δ CZA)
(iii) Prove that ar (BCZY) = ar (Δ EDZ).
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Solution
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21. In Fig. 15.84, PSDA is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove that ar (PQE) = ar (Δ CFD).
Solution
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Solution
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24. D is the mid-point of side BC of △ABC and E is the mid-point of BD. If O is the mid-point of AE, prove that ar(△BOE) = 1/8 ar(△ABC)
Solution
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25. In Fig. 15.87, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (Δ ABP) = ar (Δ ACQ).
Solution
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26. In Fig. 15.88, ABCD and AEFD are two parallelograms. Prove that
(i) PE = FQ
(ii) ar (Δ APE): ar (ΔPFA) = ar Δ (QFD) : ar (Δ PFD)
(iii) ar (Δ PEA) = ar (Δ QFD).
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Solution
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29. In figure, D and E are two points on BC such that BD = DE = EC. Show that:
ar(ΔABD)=ar(ΔADE)=ar(ΔAEC).
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Solution
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(i) ΔMBC ≅ Δ ABD
(ii) ar (BYXD) = 2 ar (Δ MBC)
(iii) ar (BYXD) = ar (ABMN)
(iv) Δ FCB ≅ Δ ACE
(v) ar (CYXE) = 2 ar (ΔFCB)
(vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar (ACFG)
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Solution
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Solution
2. In Fig. 15.75, PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of Δ OTS if PQ = 8 cm.
Solution
3. Compute the area of trapezium PQRS in figure
Solution
4. In Fig. 15.77, ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Δ AOB.
Solution
5. In Fig. 15.78, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.
Solution
6. In Fig. 15.79, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 2√5, find the area of the rectangle.
Solution
7. In Fig. 15.80, ABCD is a trapezium in which AB || DC. Prove that:
ar(Δ AOD) = ar(Δ BOC)
8. In Fig. 15.81, ABCD, ABFE and CDEF are parallelograms. Prove that
ar(Δ ADE) = ar(Δ BCF)
Solution
9. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(Δ APB) ✕ ar (Δ CPD) = ar (Δ APD) ✕ ar (Δ BPC)
Solution
10. In Fig. 15.82, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar (Δ ABC) = ar (Δ ABD)
Solution
11. If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.
Solution
12. If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of median AD, prove that ar(Δ BGC)= 2 ar (Δ AGC).
Solution
13. A point D is taken on the side BC of a ΔABC such that BD = 2DC. Prove that ar (Δ ABD) = 2 ar (Δ ADC).
Solution
14. ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that:
(i) ar (Δ ADO) = ar(Δ CDO)
(ii) ar (Δ ABP) = ar (Δ CBP).
Solution
15. ABCD is a parallelogram in which BC is produced to E such that CE= BC. AE intersects CD at F.
(i) Prove that ar(△ADF) = ar(△ECF)
(ii) If the area of △DFB = 3cm2, find the area of ∥gm ABCD.
16. ABCD is a parallelogram whose diagonals AC and BD intersect at O . A line through O intersects AB at P and DC at Q. Prove that ar(ΔPOA)=ar(ΔQOC).
Solution
17. ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.
Solution
18. In a △ABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point of AP. Prove that:
(i) ar(△PBQ) = ar(△ARC)
(ii) ar(△PRQ) = 1/2 ar(△ARC)
(iii) ar(△RQC) = 3/8 ar(△ABC)
Solution
19. ABCD is a paralleogram, G is the point on AB such that AG = 2 GB, E is a point of DC such that CE = 2 DE and F is the point of BC such that BF = 2 FC. Prove that:
(i) ar(ADEG) = ar(GBCE)
(ii) ar (△EGB) = 1/6 ar(ABCD)
(iii) ar(△EFC) = 1/2 ar(△EBF)
(iv) ar(△EBG) = ar(△EFC)
(v) Find what portion of the area of the parallelogram is the area of △EFG.
Solution
20. In Fig. 15.83, CD || AE and CY || BA.
(i) Name a triangle equal in area of ΔCBX
(ii) Prove that ar (Δ ZDE) = ar (Δ CZA)
(iii) Prove that ar (BCZY) = ar (Δ EDZ).
Solution
21. In Fig. 15.84, PSDA is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove that ar (PQE) = ar (Δ CFD).
Solution
22. In Fig. 15.85, ABCD is a trapezium in which AB∥DC and DC = 40 cm and AB = 60 cm. If X and Y are respectively the mid-points of AD and BC, prove that:
(i) XY = 50 cm
(ii) DCYX is a trapezium.
(iii) ar (trap. DCXY) = 9/11 ar (trap. XYBA)
Solution
23. In Fig. 15.86, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intesects BC in F. Prove that:
(i) ar(△BDE) = 1/4 ar(△ABC)
(ii) ar(△BDE) = 1/2 ar(△BAE)
(iii) ar(△BFE) = ar(△AFD)
(iv) ar(△ABC) = 2 ar(△BEC)
(v) ar(△FED) = 1/8 ar(△AFC)
(vi) ar(△BFE) = 2 ar(△EFD)
Solution
24. D is the mid-point of side BC of △ABC and E is the mid-point of BD. If O is the mid-point of AE, prove that ar(△BOE) = 1/8 ar(△ABC)
Solution
25. In Fig. 15.87, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (Δ ABP) = ar (Δ ACQ).
Solution
26. In Fig. 15.88, ABCD and AEFD are two parallelograms. Prove that
(i) PE = FQ
(ii) ar (Δ APE): ar (ΔPFA) = ar Δ (QFD) : ar (Δ PFD)
(iii) ar (Δ PEA) = ar (Δ QFD).
Solution
28. In a △ABC, if L and M are points on AB and AC respectively such that LM∥BC. Prove that:
(i) ar(△LCM) = ar(△LBM)
(ii) ar(△LBC) = ar(△MBC)
(iii) ar(△ABM) = ar(△ACL)
(ii) ar(△LOB) = ar(△MOC)
Solution
29. In figure, D and E are two points on BC such that BD = DE = EC. Show that:
ar(ΔABD)=ar(ΔADE)=ar(ΔAEC).
Solution
30 . If Fig. 15.91, ABC is a right triangle right angled at A, BCED, ACFGand ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y.
Show that(i) ΔMBC ≅ Δ ABD
(ii) ar (BYXD) = 2 ar (Δ MBC)
(iii) ar (BYXD) = ar (ABMN)
(iv) Δ FCB ≅ Δ ACE
(v) ar (CYXE) = 2 ar (ΔFCB)
(vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar (ACFG)
