#### 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.

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 Î” OTS if PQ = 8 cm.

3. Compute the area of trapezium PQRS in figure

4. In Fig. 15.77, âˆ AOB = 90Â°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Î” AOB.

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.

6. In Fig. 15.79, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 2âˆš5, find the area of the rectangle.

7. In Fig. 15.80, ABCD is a trapezium in which AB || DC. Prove that:

ar(Î” AOD) = ar(Î” BOC)

8. In Fig. 15.81, ABCD, ABFE and CDEF are parallelograms. Prove that

ar(Î” ADE) = ar(Î” BCF)

9. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:

ar(Î” APB) âœ• ar (Î” CPD) = ar (Î” APD) âœ• ar (Î” BPC)

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 (Î” ABC) = ar (Î” ABD)

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.

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(Î” BGC)= 2 ar (Î” AGC).

13. A point D is taken on the side BC of a Î”ABC such that BD = 2DC. Prove that ar (Î” ABD) = 2 ar (Î” ADC).

14. ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that:

(i) ar (Î” ADO) = ar(Î” CDO)

(ii) ar (Î” ABP) = ar (Î” CBP).

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(Î”POA)=ar(Î”QOC).

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.

20. In Fig. 15.83, CD || AE and CY || BA.

(i) Name a triangle equal in area of Î”CBX

(ii) Prove that ar (Î” ZDE) = ar (Î” CZA)

(iii) Prove that ar (BCZY) = ar (Î” EDZ).

24. D is the mid-point of side BC of â–³ABC and E is the mid-point of BD. If O is the mid-point of AE, prove that ar(â–³BOE) = 1/8 ar(â–³ABC)

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 (Î” ABP) = ar (Î” ACQ).

(i) PE = FQ

(ii) ar (Î” APE): ar (Î”PFA) = ar Î” (QFD) : ar (Î” PFD)

(iii) ar (Î” PEA) = ar (Î” QFD).

ar(Î”ABD)=ar(Î”ADE)=ar(Î”AEC).

(i) Î”MBC â‰… Î” ABD

(ii) ar (BYXD) = 2 ar (Î” MBC)

(iii) ar (BYXD) = ar (ABMN)

(iv) Î” FCB â‰… Î” ACE

(v) ar (CYXE) = 2 ar (Î”FCB)

(vi) ar (CYXE) = ar (ACFG)

(vii) ar (BCED) = ar (ABMN) + ar (ACFG)

**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 Î” OTS if PQ = 8 cm.

**Solution**3. Compute the area of trapezium PQRS in figure

**Solution**4. In Fig. 15.77, âˆ AOB = 90Â°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Î” 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 = 2âˆš5, find the area of the rectangle.

**Solution**7. In Fig. 15.80, ABCD is a trapezium in which AB || DC. Prove that:

ar(Î” AOD) = ar(Î” BOC)

**Solution**8. In Fig. 15.81, ABCD, ABFE and CDEF are parallelograms. Prove that

ar(Î” ADE) = ar(Î” BCF)

**Solution**9. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:

ar(Î” APB) âœ• ar (Î” CPD) = ar (Î” APD) âœ• ar (Î” 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 (Î” ABC) = ar (Î” 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(Î” BGC)= 2 ar (Î” AGC).

**Solution**13. A point D is taken on the side BC of a Î”ABC such that BD = 2DC. Prove that ar (Î” ABD) = 2 ar (Î” ADC).

**Solution**14. ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that:

(i) ar (Î” ADO) = ar(Î” CDO)

(ii) ar (Î” ABP) = ar (Î” 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(â–³ADF) = ar(â–³ECF)

(ii) If the area of â–³DFB = 3cm2, find the area of âˆ¥gm ABCD.

**Solution**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(Î”POA)=ar(Î”QOC).

**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 â–³ABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point of AP. Prove that:

(i) ar(â–³PBQ) = ar(â–³ARC)

(ii) ar(â–³PRQ) = 1/2 ar(â–³ARC)

(iii) ar(â–³RQC) = 3/8 ar(â–³ABC)

**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 (â–³EGB) = 1/6 ar(ABCD)

(iii) ar(â–³EFC) = 1/2 ar(â–³EBF)

(iv) ar(â–³EBG) = ar(â–³EFC)

(v) Find what portion of the area of the parallelogram is the area of â–³EFG.

**Solution**20. In Fig. 15.83, CD || AE and CY || BA.

(i) Name a triangle equal in area of Î”CBX

(ii) Prove that ar (Î” ZDE) = ar (Î” CZA)

(iii) Prove that ar (BCZY) = ar (Î” 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 (Î” CFD).**

**Solution**

22. In Fig. 15.85, ABCD is a trapezium in which ABâˆ¥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(â–³BDE) = 1/4 ar(â–³ABC)

(ii) ar(â–³BDE) = 1/2 ar(â–³BAE)

(iii) ar(â–³BFE) = ar(â–³AFD)

(iv) ar(â–³ABC) = 2 ar(â–³BEC)

(v) ar(â–³FED) = 1/8 ar(â–³AFC)

(vi) ar(â–³BFE) = 2 ar(â–³EFD)

**Solution**

24. D is the mid-point of side BC of â–³ABC and E is the mid-point of BD. If O is the mid-point of AE, prove that ar(â–³BOE) = 1/8 ar(â–³ABC)

**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 (Î” ABP) = ar (Î” ACQ).

**Solution****26. In Fig. 15.88, ABCD and AEFD are two parallelograms. Prove that**

(i) PE = FQ

(ii) ar (Î” APE): ar (Î”PFA) = ar Î” (QFD) : ar (Î” PFD)

(iii) ar (Î” PEA) = ar (Î” QFD).

**Solution**
28. In a â–³ABC, if L and M are points on AB and AC respectively such that LMâˆ¥BC. Prove that:

(i) ar(â–³LCM) = ar(â–³LBM)

(ii) ar(â–³LBC) = ar(â–³MBC)

(iii) ar(â–³ABM) = ar(â–³ACL)

(ii) ar(â–³LOB) = ar(â–³MOC)

**Solution**

**29. In figure, D and E are two points on BC such that BD = DE = EC. Show that:**

ar(Î”ABD)=ar(Î”ADE)=ar(Î”AEC).

**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 âŠ¥ DE meets BC at Y.

Show that(i) Î”MBC â‰… Î” ABD

(ii) ar (BYXD) = 2 ar (Î” MBC)

(iii) ar (BYXD) = ar (ABMN)

(iv) Î” FCB â‰… Î” ACE

(v) ar (CYXE) = 2 ar (Î”FCB)

(vi) ar (CYXE) = ar (ACFG)

(vii) ar (BCED) = ar (ABMN) + ar (ACFG)

**Solution**