Find distances PQ and QR.

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Q.
A garden is in the shape of a square. The gardener grew saplings of Ashoka tree on the boundary of the garden at a distance of 1 m from each other. He wants to decorate the garden with rose plants. He chose a triangular region inside the garden to grow them.

In the above situation, the gardener took help from class 10 students. They made a chart for it, which looks like the given figure

Based on the above information, answer the following questions:

(ii) (a) Find distances PQ and QR.

1 $P Q = \sqrt{17} Q R = 3 \sqrt{2}$

2 $P Q = \sqrt{16} Q R = - 3 \sqrt{2}$

3 $P Q = \sqrt{17} Q R = 3 \sqrt{4}$

4 $P Q = - \sqrt{17} Q R = 1 \sqrt{2}$

Ans.
(1)

We have P(4, 6), Q(3, 2) and R(6, 5)

Finding distances PQ and QR

$P Q = \sqrt{(3 - 4)^{2} + (2 - 6)^{2}} = \sqrt{1 + 16} = \sqrt{17} R = \sqrt{(6 - 3)^{2} + (5 - 2)^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2}$

Solution

1 $P Q = \sqrt{17} Q R = 3 \sqrt{2}$

2 $P Q = \sqrt{16} Q R = - 3 \sqrt{2}$

3 $P Q = \sqrt{17} Q R = 3 \sqrt{4}$

4 $P Q = - \sqrt{17} Q R = 1 \sqrt{2}$

Ans.
(1)

We have P(4, 6), Q(3, 2) and R(6, 5)

Finding distances PQ and QR

$P Q = \sqrt{(3 - 4)^{2} + (2 - 6)^{2}} = \sqrt{1 + 16} = \sqrt{17} R = \sqrt{(6 - 3)^{2} + (5 - 2)^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2}$

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Solution

1 PQ = 17QR = 32

2 PQ = 16QR = -32

3 PQ = 17QR = 34

4 PQ = -17QR = 12

Ans. (1) We have P(4, 6), Q(3, 2) and R(6, 5) Finding distances PQ and QR PQ=(3-4)2+(2-6)2=1+16=17R=(6-3)2+(5-2)2=9+9=18=32

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1 $P Q = \sqrt{17} Q R = 3 \sqrt{2}$

2 $P Q = \sqrt{16} Q R = - 3 \sqrt{2}$

3 $P Q = \sqrt{17} Q R = 3 \sqrt{4}$

4 $P Q = - \sqrt{17} Q R = 1 \sqrt{2}$

Ans.
(1)

We have P(4, 6), Q(3, 2) and R(6, 5)

Finding distances PQ and QR

$P Q = \sqrt{(3 - 4)^{2} + (2 - 6)^{2}} = \sqrt{1 + 16} = \sqrt{17} R = \sqrt{(6 - 3)^{2} + (5 - 2)^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2}$