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In Figure-2, DE \parallel BC . If  \frac{AD}{DB}= \frac{3}{2}  and AE = 2.7cm then EC is equal to
Option: 1 2.0 cm 
Option: 2 1.8cm 
Option: 3 4.0 cm 
Option: 4 2.7 cm 

By using the proportionality theorem, we get

     \frac{AD}{DB}=\frac{AE}{EC}

\\\frac{3}{2}=\frac{2.7}{EC}\\EC=2.7\times\frac{2}{3}=1.8cm

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In Figure-1, ABC is an isosceles triangle, right-angled at C. Therefore 
Option: 1 AB^{2}=2AC^{2}
Option: 2 BC^{2}=2AB^{2}
Option: 3 AC^{2}=2AB^{2}
Option: 4 AB^{2}=4AC^{2}

\\ AB^2 = AC^2+BC^2 \\ AB^2 = AC^2+AC^2 \because AC=BC\\ AB^2 = 2AC^2

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It is being given that the points A\left ( 1,2 \right ), B\left ( 0,0 \right ) and C\left ( a,b \right ) are collinear. Which of the following relations between a and b is true ? 
Option: 1 a = 2b
Option: 2 2a=b
Option: 3 a+b = 0
Option: 4 a-b=0

\\$Area of triangle $ = \frac{1}{2}[1(0-\mathrm{b})-2(0-\mathrm{a})+1(0-0)] \\ A, B, C$ are collinear, the area of the triangle $\Delta A B C is zero. \\ \frac{1}{2}[1(0-\mathrm{b})-2(0-\mathrm{a})+1(0-0)]=0 \\ -\mathrm{b}+2 \mathrm{a}=0$ \\ $2 a=b$ \\

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If  \small \DeltaABC ~ \small \DeltaDEF such that AB = 1.2 cm and DE = 1.4 cm, the ratio of the areas of \small \DeltaABC and \small \DeltaDEF is  
Option: 1 49:36
Option: 2 6:7
Option: 3 7:6
Option: 4 36:49

The ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

Therefore the ratio of areas is 

\frac{area(\Delta ABC)}{area(\Delta DEF)}=(\frac{1.2}{1.4})^2=\frac{36}{49}

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ABC is an equilateral triangle of side 2a, then length of one of its altitude is ________.
 

\\\text{Length of the altitude is}=\sqrt{(2a)^2-a^2}\\=\sqrt{4a^2-a^2}=\sqrt{3}a

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Given \Delta ABC\sim \Delta PQR, if \frac{AB}{PQ}=\frac{1}{3}, then \frac{ar(\Delta ABC)}{ar(\Delta PQR)}=-________. 
 

\frac{area(\Delta ABC)}{area(\Delta PQR)}=\frac{1}{9}

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The ratio of the corresponding altitudes of two similar triangles is 3/5. Is it correct to say that ratio of their areas is 6/5? Why?

Answer: [False]

Given:- Ratio of corresponding altitudes of two similar triangles is 3/5

As we know that the ratio of the areas of two similar triangles is equal to the ratio of squares of any two corresponding altitudes.

\therefore \frac{\text {Area 1}}{\text {Area 2}}=\left ( \frac{\text {Altitude 1}}{\text {Altitude 2}} \right )^{2}

\frac{\text {Area 1}}{\text {Area 2}}=\left ( \frac{3}{5} \right )^{2}            \left [ Q\frac{\text {Altitude 1}}{\text {Altitude 2}}=\frac{3}{5} \right ]

=\frac{9}{25}

Hence the given statement is false because ratio of areas of two triangles is 9/25 which is not equal to 6/5

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If in two right triangles, one of the acute angles of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles will be similar? Why?

Answer : [True]

Let two right angle triangles are ABC and PQR.

Given:- One of the acute angles of one triangle is equal to an acute angle of the other triangle.

\text {In}\Delta ABC \; \text {and}\; \Delta PQR

\\\angle B = \angle Q = 90^{o}\\\angle C = \angle R = x^{o}                         

In \Delta ABC

\angle A + \angle B + \angle C = 180^{o}          

[Sum of interior angles of a triangle is 180°]

\angle A + 90^{o}+ x^{o} = 180^{o}

\angle A + 90^{o}- x^{o} \; \; \; \; \; \; \; \; \; ....(1)

Also in \Delta PQR

\angle P + \angle Q + \angle R = 180^{o}

\angle P + 90^{o}- x^{o} = 180^{o}

\angle P = 90^{o}- x^{o} \; \; \; \; \; \; .....(2)

from equation (1) and (2)

\angle A=\angle P \; \; \; \; \; \; .....(3)

From equations (1), (2), and (3) are observed that the corresponding angles of the triangles are equal therefore triangles are similar.

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D is a point on side QR of \DeltaPQR such that PD \perp QR. Will it be correct to say that \DeltaPQD ~ \DeltaRPD? Why?

Answer : [False]

In \DeltaPQD and \DeltaRPD

PD = PD          (common side)

\angle PDQ = \angle PDR            (each 90o)

The given statement \Delta PQD \sim \Delta RPD is false.

Because according to the definition of similarity, two triangles are similar, if their corresponding angles are equal and their corresponding sides are in the same ratio

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In Fig., if \angleD = \angleC, then is it true that \Delta ADE \sim \Delta ACB ? Why?

Answer : [True]

In \Delta ADE and \Delta ACB

\angle D=\angle C        (given)

\angle A = \angle A      (common angle)

And we know that if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criteria.

\therefore   \Delta ADE \sim \Delta ACB         {by AA similarity criterion}

Hence the given statement is true.

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