# FRANK Solutions for Class 9 Maths Chapter 21 - Areas Theorems on Parallelograms

Get quick access to online Frank Solutions for ICSE Class 9 Mathematics Chapter 21 Areas theorems on Parallelograms on TopperLearning. You will learn how to prove theorems in this chapter. You can use the textbook solutions as quick reference materials to revise the lessons.

Learn to use the accurate theorem to find the area of a parallelogram using the information of area of a triangle that’s part of the parallelogram. Revise similar area theorem problems effectively with our ICSE Class 9 Maths Frank solutions. For clarity on quadrilateral problems, you can get your doubts answered in our learning portal’s ‘UnDoubt’ section.

## Chapter 21 - Areas Theorems on Parallelograms Exercise Ex. 21.1

In
the given figure area of ∥ gm PQRS is 30 cm^{2}. Find the height of ∥
gm PQFE if PQ = 6 cm.

In the given figure, PQRS is a ∥ gm. A straight line through P cuts SR at point T and QR produced at N. Prove that area of triangle QTR is equal to the area of triangle STN.

ΔPQT and parallelogram PQRS are on the same base PQ and between the same parallel lines PQ and SR.

ΔPSN and parallelogram PQRS are on the same base PS and between the same parallel lines PS and QN.

Adding equations (i) and (ii), we get

Subtracting A(ΔRTN) from both the sides, we get

In the given figure, ST ∥ PR. Prove that: area of quadrilateral PQRS = area of ΔPQT.

In the given figure, if AB ∥ DC ∥ FG and AE is a straight line. Also, AD ∥ FC. Prove that: area of ∥ gm ABCD = area of ∥ gm BFGE.

* *

Joining AC and FE, we get

ΔAFC and ΔAFE are on the same base AF and between the same parallels AF and CE.

⇒ A(ΔAFC) = A(ΔAFE)

⇒ A(ΔABF) + A(ΔABC) = A(ΔABF) + A(ΔBFE)

⇒ A(ΔABC) = A(ΔBFE)

⇒ A(parallelogram ABCD) = A(parallelogram BFGE) ⇒ A(parallelogram ABCD) = A(parallelogram BFGE)

In the given figure, the perimeter of parallelogram PQRS is 42 cm. Find the lengths of PQ and PS.

In the given figure, PT ∥ QR and QT ∥ RS. Show that: area of ΔPQR = area of ΔTQS.

*Question modified

Joining TR, we get

ΔPQR and ΔQTR are on the same base QR and between the same parallel lines QR and PT.

ΔQTR and ΔTQS are on the same base QT and between the same parallel lines QT and RS.

From (i) and (ii), we get

In the given figure, ΔPQR is right-angled at P. PABQ and QRST are squares on the side PQ and hypotenuse QR. If PN ⊥ TS, show that:

(a) ΔQRB ≅ ΔPQT

(b) Area of square PABQ = area of rectangle QTNM.

The diagonals of a parallelogram ABCD intersect at O. A line through O meets AB in P and CD in Q. Show that

In the given figure AF = BF and DCBF is a parallelogram. If the area of ΔABC is 30 square units, find the area of the parallelogram DCBF.

In the given figure, BC ∥ DE.

(a) If area of ΔADC is 20 sq. units, find the area of ΔAEB.

(b) If the area of ΔBFD is 8 square units, find the area of ΔCEF

In the given figure, AB ∥ SQ ∥ DC and AD ∥ PR ∥ BC. If the area of quadrilateral ABCD is 24 square units, find the area of quadrilateral PQRS.

In the given figure, PQ ∥ SR ∥ MN, PS ∥ QM and SM ∥ PN. Prove that:

ar. (SMNT) = ar. (PQRS).

SM ∥ PN

⇒ SM ∥ TN

Also, SR ∥ MN

⇒ ST ∥ MN

Hence, SMNT is a parallelogram.

SM ∥ PN

⇒ SM ∥ PO

Also, PS ∥ QM

⇒ PS ∥ OM

Hence, SMOP is a parallelogram.

Now, parallelograms SMNT and SMOP are on the same base SM and between the same parallels SM and PN.

∴ A(parallelogram SMNT) = A(parallelogram SMOP) ….(i)

Similarly, we can show that quadrilaterals PQRS is a parallelogram.

Now, parallelograms PQRS and SMOP are on the same base PS and between the same parallels PS and QM.

∴ A(parallelogram PQRS) = A(parallelogram SMOP) ….(ii)

From (i) and (ii), we have

A(parallelogram SMNT) = A(parallelogram PQRS)

In the given figure, ABC is a triangle and AD is the median.

If E is any point on the median AD. Show that:

Area of ΔABE = Area of ΔACE.

AD is the median of ΔABC.

Therefore it will divide ΔABC into two triangles of equal areas.

∴ Area(ΔABD) = Area(ΔACD) ….(i)

Similarly, ED is the median of ΔEBC.

∴ Area(DEBD) = Area(DECD) ….(ii)

Subtracting equation (ii) from (i), we have

Area(ΔABD) - Area(ΔEBD) = Area(ΔACD) - Area(ΔECD)

⇒ Area(ΔABE) = Area(ΔACE)

In the given figure, ABC is a triangle and AD is the median.

If E is the midpoint of the median AD, prove that:

Area of ΔABC = 4 × Area of ΔABE

AD is the median of ΔABC.

Therefore it will divide ΔABC into two triangles of equal areas.

∴ Area(ΔABD) = Area(ΔACD) ….(i)

Similarly, ED is the median of ΔEBC.

∴ Area(ΔEBD) = Area(ΔECD) ….(ii)

Subtracting equation (ii) from (i), we have

Area(ΔABD) - Area(ΔEBD) = Area(ΔACD) - Area(ΔECD)

⇒ Area(ΔABE) = Area(ΔACE) ….(iii)

Since E is the mid-point of median AD,

AE = ED

Now,

ΔABE and ΔBED have equal bases and a common vertex B.

∴ Area(ΔABE) = Area(ΔBED) ….(iv)

From (i), (ii), (iii) and (iv), we get

Area(ΔABE) = A(ΔBED) = Area(ΔACE) = Area(ΔEDC) ….(v)

Now,

Area(ΔABC) = Area(ΔABE) + A(ΔBED) + Area(ΔACE) + Area(ΔEDC)

= 4 × Area(ΔABE) [From (v)]

In a parallelogram PQRS, M and N are the midpoints of the sides PQ and PS respectively. If area of ΔPMN is 20 square units, find the area of the parallelogram PQRS.

Construction: Join SM and SQ.

In a parallelogram PQRS, SQ is the diagonal.

So, it bisects the parallelogram.

∴ Area(DPSQ)

SM is the median of ΔPSQ.

∴ Area(ΔPSM)

Again, MN is the median of ΔPSM.

∴ Area(ΔPMN)

In a parallelogram PQRS, T is any point on the diagonal PR. If the area of ΔPTQ is 18 square units find the area of ΔPTS.

Construction: Join QR. Let the diagonals PR and QS intersect each other at point O.

Since diagonals of a parallelogram bisect each other, therefore O is the mid-point of both PR and QS.

Now, median of a triangle divides it into two triangles of equal area.

In ΔPSQ, OP is the median.

∴ Area(ΔPOS) = Area(ΔPOQ) ….(i)

Similarly, OT is the median of ΔTSQ.

∴Area(ΔTOS) = Area(ΔTOQ) ….(ii)

Subtracting equation (ii) from (i), we have

Area(ΔPOS) - Area(ΔTOS) = Area(ΔPOQ) - Area(ΔTOQ)

⇒ Area(ΔPTQ) = Area(ΔPTS)

⇒ Area(ΔPTS) = 18 square units

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