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Standard X

Mathematics

Difference between Congruency and Similarity

Question

In Figure 3, PQ || CD and PR || CB. Prove that $\frac{AQ}{QD} = \frac{AR}{RB}$ .

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Solution

Given that:

and , then we to prove that

The following diagram is given

We can easily see that, in the above figure are similar triangles, and also the are similar triangles.

Now, we have the following properties of similar triangles,

From equation (1) and Equation (2), we get

Hence proved.

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Similar questions

Find the area of the given figure if each grid occupies 1 sq cm.

If $c o s (α + β) s i n (γ + δ) = c o s (α - β) s i n (γ - δ), p r o v e t h a t c o t α c o t β c o t γ = c o t δ$

Δ

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\sim Δ

LBN, ln

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∠

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. Construct

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Q. Prove that

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$= 3 sin (α - β) sin (β - γ) sin (γ - δ) sin (α - δ) sin (β - δ) sin (γ - δ)$

Q. Given

Δ

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\sim

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\frac{A B}{P Q} = \frac{1}{3}

, then find

\frac{a r Δ A B C}{a r Δ P Q R}

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In Figure 3, PQ || CD and PR || CB. Prove that AQQD=ARRB.

Given that: and , then we to prove that The following diagram is given We can easily see that, in the above figure are similar triangles, and also the are similar triangles. Now, we have the following properties of similar triangles, From equation (1) and Equation (2), we get Hence proved.

In Figure 3, PQ || CD and PR || CB. Prove that $\frac{AQ}{QD} = \frac{AR}{RB}$ .

Given that:

and , then we to prove that

The following diagram is given

We can easily see that, in the above figure are similar triangles, and also the are similar triangles.

Now, we have the following properties of similar triangles,

From equation (1) and Equation (2), we get

Hence proved.