5. In the right angled ∆PQR, m P = 90°. If I(PQ) = 15 cm, and I (PR) = 8 cm,
find the length of seg QR.
Answers
Answer:
Given:
PQ = 8 cm
QR = 6 cm
PR = ?
∠PQR = 90°
According to Pythagoras Theorem,
PR^2=PQ^2+QR^2PR
2 =PQ
2 +QR
2
PR^2=8^2+6^2PR
2 =8
2 +62
PR^2=64+36PR
2=64+36
PR^2=100PR
2=100
\therefore PR=\sqrt{100}=10\ cm∴PR=
100 =10 cm
(ii) Given :
PR = 34 cm
QR = 30 cm
PQ = ?
∠PQR = 90°
Answer:
Given:
PQ = 8 cm
QR = 6 cm
PR = ?
∠PQR = 90°
According to Pythagoras Theorem,
PR^2=PQ^2+QR^2PR
2
=PQ
2
+QR
2
PR^2=8^2+6^2PR
2
=8
2
+6
2
PR^2=64+36PR
2
=64+36
PR^2=100PR
2
=100
\therefore PR=\sqrt{100}=10\ cm∴PR=
100
=10 cm
(ii) Given :
PR = 34 cm
QR = 30 cm
PQ = ?
∠PQR = 90°
According to Pythagoras Theorem,
PR^2=PQ^2+QR^2PR
2
=PQ
2
+QR
2
34^2=PQ^2+30^234
2
=PQ
2
+30
2
1156=PQ^2+9001156=PQ
2
+900
1156-900=PQ^21156−900=PQ
2
256=PQ^2256=PQ
2
\therefore PQ=16\ cm∴PQ=16 cm
Step-by-step explanation:
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