**Exercise 15.1**

**Question 1: Which of the following figures lie on the same base and between the same parallel. In such a case, write the common base and two parallels:**

**Solution:**

**Exercise 15.2**

**Question 1: If figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.**

**Solution:**

**Question 2: In Q.No. 1, if AD = 6 cm, CF = 10 cm and AE = 8 cm, find AB.**

**Solution:**

**Question 3: Let ABCD be a parallelogram of area 124 cm2. If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AEFD.**

**Solution:**

^{2}

^{2}

^{2}.

**Question 4: If ABCD is a parallelogram, then prove that ar(ΔABD) = ar(Δ BCD) = ar(Δ ABC) = ar(Δ ACD) = 1/2 ar(||gm ABCD)**

**Solution:**

**Exercise 15.3**

**Question 1: In figure, compute the area of quadrilateral ABCD.**

**Solution:**

^{2}= BD

^{2}+ BC

^{2}(By Pythagorean Theorem)

^{2}= BD

^{2}+ 8

^{2}

^{2}+ 64

^{2}= 289 – 64

^{2}= 225

^{2}= AB

^{2}+ AD

^{2}(By Pythagorean Theorem)

^{2}= AB

^{2}+ 9

^{2}

^{2}+ 9

^{2}

^{2 }= 225−81

^{2}= 144

^{2}

^{2}

^{2 }

^{2}

^{2}+ 68 cm

^{2}

^{2}

^{2}.

**Question 2: In figure, PQRS is a square and T and U are, respectively, the mid-points of PS and QR . Find the area of ΔOTS if PQ = 8 cm.**

**Solution:**

^{2}

^{2}

^{2}.

**Question 3: Compute the area of trapezium PQRS in figure.**

**Solution:**

^{2}= QT

^{2}+ RT

^{2}(Using Pythagorean Theorem)

^{2}= QR

^{2}− QT

^{2}

^{2}= 172 − 82

^{2}= 225

^{2 }

^{2}

^{2}.

**Question 4: In figure, ∠AOB = 90**

^{o}, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of ΔAOB.**Solution:**

^{2}= OB

^{2}+ OA

^{2}(Using Pythagorean Theorem)

^{2}= OB

^{2}+ 12

^{2}

^{2}= 169 – 144

^{2}= 25

^{2 }

^{2}

^{2}

^{2}.

**Question 5: In figure, 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:**

^{2}= BM

^{2}+ MC

^{2}(Pythagorean Theorem)

^{2}

^{2}

^{2}= 25 – 16

^{2 }= 9

^{2}= AL

^{2}+ DL

^{2}(Pythagorean Theorem)

^{2}= 4

^{2}+ DL

^{2}

^{2}

^{2}= 25 – 16 = 9

^{2}.

**Question 6: In figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 2√5 cm, find the area of the rectangle.**

**Solution:**

^{2}= OE

^{2}+ DE

^{2}

^{2}= (2√5 )

^{2}+ DE

^{2}

^{2}

^{2}.

**Question 7: In figure, ABCD is a trapezium in which AB || DC. Prove that ar(ΔAOD) = ar(ΔBOC)**

**Solution:**

**Question 8: In figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(ΔADE) = ar(ΔBCF).**

**Solution:**

**Question 9: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar(ΔAPB) × ar(ΔCPD) = ar(ΔAPD) × ar(ΔBPC).**

**Solution:**

**Question 10: In figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar(ΔABC) = ar(ΔABD).**

**Solution:**

**Question 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:**

**Question 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 the median AD, prove that ar(ΔBGC) = 2ar(ΔAGC).**

**Solution:**

**Question 13: A point D is taken on the side BC of a ΔABC, such that BD = 2DC. Prove that ar(ΔABD) = 2ar(ΔADC).**

**Solution:**

**Question 14: ABCD is a parallelogram whose diagonal intersects at O .If P is any point on BO, prove that:**

**(i) ar(ΔADO) = ar(ΔCDO).**

**(ii) ar(ΔABP) = 2ar(ΔCBP).**

**Solution:**

**Question 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(ΔADF) = ar(ΔECF).**

**(ii) If the area of ΔDFB = 3 cm**

^{2}, find the area of ∥gm ABCD.**Solution:**

^{2 }

^{2}

^{2 }

^{2}

^{2}.

**Question 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:**

**Question 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:**

**Question 18: In a triangle ABC, P and Q are respectively the mid points of AB and BC and R is the midpoint of AP. Prove that:**

**(i) ar(ΔPBQ) = ar(ΔARC).**

**(ii) ar(ΔPRQ) = 1/2ar(ΔARC).**

**(iii) ar(ΔRQC) = 3/8ar(ΔABC).**

**Solution:**

**Question 19: ABCD is a parallelogram. G is a point on AB such that AG = 2GB and E is point on DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:**

**(i) ar(ADEG) = ar(GBCE)**

**(ii) ar(ΔEGB) = 1/6 ar(ABCD)**

**(iii) ar(ΔEFC) = 1/2 ar(ΔEBF)**

**(iv) ar(ΔEGB) = 3/2 × ar(ΔEFC)**

**(v) Find what portion of the area of parallelogram is the area of ΔEFG.**

**Solution:**

**Question 20: In figure, CD ∥ AE and CY ∥ BA.**

**(i) Name a triangle equal in area of ΔCBX**

**(ii) Prove that ar(ΔZDE) = ar(ΔCZA)**

**(iii) Prove that ar(BCZY) = ar(ΔEDZ)**

**Solution:**

**Question 21: In figure, PSDA is a parallelogram in which PQ = QR = RS and AP ∥ BQ ∥ CR. Prove that ar (ΔPQE) = ar(ΔCFD).**

**Solution:**

**Question 22: In figure, 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. DCYX) = 9/11 ar(XYBA).**

**Solution:**

^{2 }

^{2}

^{2 }

^{2}

**Question 23: In figure ABC and BDE are two equilateral triangles such that D is the midpoint of BC. AE intersects BC in F. Prove that:**

**(i) ar(ΔBDE) = 1/4 ar(ΔABC)**

**(ii) ar(ΔBDE) = 1/2ar(ΔBAE)**

**(iii) ar(ΔBFE) = ar(ΔAFD)**

**(iv) ar(ΔABC) = 2 ar(ΔBEC)**

**(v) ar(ΔFED) = 1/8 ar(ΔAFC)**

**(vi) ar(ΔBFE) = 2 ar(ΔEFD)**

**Solution:**

^{2}

^{2}

^{2}/4)

**Question 24: D is the midpoint of side BC of ΔABC and E is the midpoint of BD. If O is the midpoint of AE, Prove that ar(ΔBOE) = 1/8 ar(ΔABC).**

**Solution:**

**Question 25: In figure, X and Y are the mid points of AC and AB respectively, QP ∥ BC and CYQ and BXP are straight lines. Prove that ar(ΔABP) = ar(ΔACQ).**

**Solution:**

**Question 26: In figure, ABCD and AEFD are two parallelograms. Prove that**

**(i) PE = FQ**

**(ii) ar(ΔAPE) : ar(ΔPFA) = ar(ΔQFD): ar(ΔPFD)**

**(iii) ar(ΔPEA) = ar(ΔQFD)**

**Solution:**

**Question 27: In figure, ABCD is a parallelogram. O is any point on AC. PQ ∥ AB and LM ∥ AD. Prove that: ar(∥gm DLOP) = ar(∥gm BMOQ).**

**Solution:**

**Question 28: In a triangle ABC, if L and M are points on AB and AC respectively such that LM ∥ BC. Prove that:**

**(i) ar(ΔLCM) = ar(ΔLBM)**

**(ii) ar(ΔLBC) = ar(ΔMBC)**

**(iii) ar(ΔABM) = ar(ΔACL)**

**(iv) ar(ΔLOB) = ar(ΔMOC)**

**Solution:**

**Question 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:**

**Question: 30 In figure, ABC is a right angled triangle at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that**

**(i) ΔMBC ≅ ΔABD**

**(ii) ar(BYXD) = 2ar(ΔMBC)**

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

**(iv) ΔFCB ≅ ΔACE**

**(v) ar(CYXE) = 2ar(ΔFCB)**

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

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

**Solution:**

^{2}= AB

^{2}+ AC

^{2}(By Pythagoras theorem in triangle ACB)

**Question 1: If ABC and BDE are two equilateral triangles such that D is the mid-point of BC, then find ar(△ABC) : ar(△BDE).**

**Solution:**

^{2}

**Question 2: In figure, ABCD is a rectangle in which CD = 6 cm, AD = 8 cm. Find the area of parallelogram CDEF.**

**Solution:**

^{2}

^{2}

**Question 3: In figure, find the area of ΔGEF.**

**Solution:**

^{2}

^{2}

^{2})

^{2}.

**Question 4: In figure, ABCD is a rectangle with sides AB = 10 cm and AD = 5 cm. Find the area of ΔEFG.**

**Solution:**

^{2}

^{2}

^{2})

^{2}.

**Question 5: PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then find ar(△RAS).**

**Solution:**

^{2}= PR

^{2}– PS

^{2}(Using Pythagoras theorem)

^{2}=(13)

^{2}– (5)

^{2}

^{2}= 169 – 25

^{2}= 114

^{2}.