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Area of a Parallelogram

Question 9In ...

Question

Question 9

In $Δ P Q R$ ,

a) PQ - QR > PR
b) PQ + QR < PR
c) PQ - QR < PR
d) PQ + PR < QR

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Solution

As we know, sum of the lengths of any two sides of a triangle is always greater than the length of the third side.

In $Δ P Q R$ , $P R + Q R > P Q$
$\Rightarrow P R > P Q - Q R \Rightarrow P Q - Q R < P R$
Hence option (c) is correct.

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Similar questions

Q. Question 9
In

Δ P Q R

∠ R > ∠ Q

, then
(A) QR > PR
(B) PQ > PR
(C) PQ < PR
(D) QR < PR

⎡ ⎢ ⎣ \begin{matrix} - q r & p (q + r) & p r + p q p q + q r & - p r & p q + q r q r + p r & q r + p r & - p q \end{matrix} ⎤ ⎥ ⎦

| A |

(p q - q r + p r) (p q + q r) - (p r + p q) (p + q - r)

Q. In ΔPQR, if

∠ R > ∠ Q

, then

(a) QR > PR

(b) PQ > PR

(c) PQ < PR

(d) QR < PR

Q. (a) Construct

Δ P Q R

¯ ¯¯¯¯¯¯ ¯ P Q = 3.5 c m, ¯ ¯¯¯¯¯¯ ¯ P R = 4.5 c m

¯ ¯¯¯¯¯¯¯ ¯ Q R = 5.5 c m

(b) Construct ΔPQR, when PQ = 4 cm, QR = 6 cm and ∠PQR = 60

^{\circ}

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Area of a Parallelogram

Standard VII Mathematics

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