Selina solutions concise maths class 9 icse Solutions for Chapter 16 chapter 16 area theorems

Exercises in Chapter 16 chapter 16 area theorems Grade 9

Exercise 16(A)

Exercise 16(B)

Exercise 16(C)

Questions in Exercise 16(A)

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 (𝚫POS)= Area (//gm PMLS)

(ii) Area (𝚫POS) + Area (𝚫QOR)= Area (//gm PQRS)

(iii)Area (𝚫POS) + Area (𝚫QOR)= Area (𝚫POQ) + Area (𝚫SOR)

In parallelogram ABCD, P is a point on side AB and Q is a point on side BC. Prove that:

(i) 𝚫COP and 𝚫AQD are equal in the area.

(ii) Area (𝚫AQD) = Area (𝚫APD) + Area (𝚫CPB)

In the following figure, CE is drawn parallel to diagonal DB of the quadrilateral ABCD which meets AB produced at point E.

Prove that triangle ADE and Quadrilateral ABCD are equal in area.

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.

In the following figure, DE is parallel to BC. Show that:

(i) Area (∆ADC) = Area (∆AEB)

(ii) Area (∆BOD) = Area (∆COE)

. ABCD and BCFE are parallelograms. If area of triangle EBC=480cm2 , AB=30cm and BC=40cm; Calculate;

(i) Area of parallelogram ABCD;

(ii) Area of parallelogram BCFE;

(iii) Length of altitude from A on CD;

(iv) Area of triangle ECF.

In the given figure, D is the mid-point of side AB of ∆ABC and BDEC is a parallelogram. Prove that:

Area of ∆ABC = Area of ||gm BDEC.

In the following figure, AC||PS||QR and PQ||DB||SR. Prove that:

Area of Quadrilateral PQRS = 2 x Area of quad. ABCD.

ABCD is a trapezium with AB||DC. A line parallel to AC intersects AB at point M and BC at point N. Prove that:

Area of triangle ADM = area of triangle ACN

In the given figure,

AD||BE||CF.

Prove that:

Area (∆AEC) = Area (∆DBF)

The given figure shows the parallelograms ABCD and APQR. Show that these parallelograms are equal in the area.

[Join B and R]

Questions in Exercise 16(B)

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 of the same base is equal to the ratio of their heights.

In the given figure; AD is the median of ∆ABC and E is any point on median AD. Prove that area of triangle ABE = area of triangle ACE.

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 parallelogram ABCD.

The base BC of triangle ABC is divided at D so that BD=1/2 DC. Prove that area of the triangle ABD=1/3 of the area of triangle ABC.

In a parallelogram ABCD, point P lies in DC such that DP: PC=3:2. If the area of triangle DPB=30 sq. cm, find the area of the parallelogram ABCD.

The following figure shows a triangle ABC in which P, Q, and R are mid-points of sides AB, BC and CA respectively. S is the mid-point of PQ.

Prove that: ar.(∆ ABC) = 8 x ar.(∆ QSB)

Questions in Exercise 16(C)

The given figure shows a parallelogram ABCD with area 324 sq. cm. P is a point in AB such that AP: PB=1:2. Find the area of ∆APD.

In parallelogram ABCD, P is the mid-point of AB. CP and BD intersect each other at point O. If the area of triangle POB = 40 cm2 and OP:OC = 1:2, find:

(i) Areas of triangle BOC and PBC

(ii) Areas of triangle ABC and parallelogram ABCD.

The medians of a triangle ABC intersect each other at point G. If one of its medians is AD, prove that:

(i) Area (∆ABD) = 3 × Area (∆BGD)

(ii) Area (∆ACD) = 3 × Area (∆CGD)

(iii) Area (∆BGC) = 1/3 × Area (∆ABC)

The perimeter of a triangle ABC is 37 cm and the ratio between the lengths of its altitudes be 6:5:4. Find the lengths of its sides.

In the following figure, OAB is a triangle and AB||DC. If the area od triangle CAD = 140 cm2 and the area of triangle ODC = 172 cm2 , find

(i) The area of triangle DBC

(ii) The area of triangle OAC

(iii) The area of triangle ODB

Was This helpful?

Chapters in this book

Chapter 1- Rational and Irrational Numbers

Chapter 2- Compound Interest [Without Using Formula

Chapter 3- Compound Interest [Using Formula

Chapter 4- Expansions

Chapter 5- Factorisation

Chapter 6- Simultaneous Equations

Chapter 7- Indices [exponents]

Chapter 8- Logarithms

Chapter 9- Triangles [Congruency in Triangles]

Chapter 10- Isosceles Triangle

Chapter 11- Inequalities

Chapter 12- Mid-Point and Its Converse

Chapter 13- Pythagoras Theorem

Chapter 14- Rectilinear Figures

Chapter 15- Construction of Polygons

Chapter 16- Area Theorems

Chapter 17- Circles

Chapter 18- Statistics

Chapter 19- Mean and Median

Chapter 20- Area and Perimeter of Plane Figures

Chapter 21- Solids

Chapter 22- Trigonometrical Ratios

Chapter 23- Trigonometrical Ratios of Standard Angles

Chapter 24- Solution Of Right Triangles

Chapter 25- Complementary angles

Chapter 26- Co-ordinate Geometry

Chapter 27- Graphical Solution

Chapter 28- Distance Formula