EduSahara™ Worksheet
Name : Chapter Based Worksheet
Chapter : Pythagoras Theorem
Grade : ICSE Grade IX
License : Non Commercial Use
Question
1
1.
In a right angled triangle △PQR, if QR = 13 cm, PQ = 19 cm are the lengths of perpendicular sides , then corresponding height of side RP =
(i)
7.73 cm
(ii)
10.73 cm
(iii)
5.73 cm
(iv)
13.73 cm
(v)
15.73 cm
Question
2
2.
In the given figure, △BDC is right-angled at D, DE ⟂ BC.
BC
= c,
DC
= a,
BD
= b and
DE
= p.
Which of the following are true?
a)
1
a
2
+
1
b
2
=
1
c
2
+
1
p
2
b)
1
a
2
+
1
b
2
=
1
p
2
c)
1
a
2
+
1
b
2
+
1
c
2
=
1
p
2
d)
ab
=
pc
e)
a
2
+
b
2
=
c
2
(i)
{c,d}
(ii)
{b,d,e}
(iii)
{a,b}
(iv)
{a,c,e}
(v)
{a,b,d}
Question
3
3.
A vehicle goes 10 km South and then 12 km West. How far is it from its starting point ?
(i)
16.62 km
(ii)
17.62 km
(iii)
14.62 km
(iv)
15.62 km
(v)
13.62 km
Question
4
4.
In a right angled triangle, if the two non-hypotenuse sides are 10 cm and 24 cm, find the hypotenuse
(i)
25.00 cm
(ii)
28.00 cm
(iii)
27.00 cm
(iv)
24.00 cm
(v)
26.00 cm
Question
5
5.
In a right angled triangle △PQR, if the area is 112 sq.cm and base QR = 14 cm, then side RP =
(i)
26.26 cm
(ii)
16.26 cm
(iii)
21.26 cm
(iv)
18.26 cm
(v)
24.26 cm
Question
6
6.
In the given figure, △DEF is an acute angled triangle and DG ⟂ EF. Then
(i)
DF
2
=
DE
2
+
EF
2
+
2
EF
.
EG
(ii)
DF
2
=
DE
2
+
EF
2
−
DG
2
(iii)
DF
2
=
DE
2
+
EF
2
−
2
DE
.
EF
(iv)
DF
2
=
DE
2
+
EF
2
−
2
EF
.
EG
(v)
DF
2
=
DE
2
+
EF
2
+
2
DE
.
EF
Question
7
7.
In an isosceles right angled triangle △PQR, if perimeter = 47.8 cm, then side RP =
(i)
22.80 cm
(ii)
19.80 cm
(iii)
16.80 cm
(iv)
24.80 cm
(v)
14.80 cm
Question
8
8.
In an isosceles right angled triangle △PQR, if QR = 17 cm is one of the equal sides, then perimeter of the triangle =
(i)
61.04 cm
(ii)
55.04 cm
(iii)
63.04 cm
(iv)
58.04 cm
(v)
53.04 cm
Question
9
9.
In a right angled triangle △PQR, if QR = 19 cm is one of the perpendicular sides and RP = 25.5 cm is the hypotenuse, then area of the triangle =
(i)
189.50 sq.cm
(ii)
137.50 sq.cm
(iii)
173.50 sq.cm
(iv)
148.50 sq.cm
(v)
161.50 sq.cm
Question
10
10.
In a right angled triangle △PQR, if the area is 140 sq.cm and corresponding height of side QR = 14 cm, then side RP =
(i)
19.41 cm
(ii)
24.41 cm
(iii)
21.41 cm
(iv)
29.41 cm
(v)
27.41 cm
Question
11
11.
In an isosceles right angled triangle △PQR, if corresponding height to the base QR is 19 cm, then side PQ =
(i)
24.00 cm
(ii)
19.00 cm
(iii)
16.00 cm
(iv)
22.00 cm
(v)
14.00 cm
Question
12
12.
In an isosceles right angled triangle △PQR, if QR = 11 cm is one of the equal sides, then corresponding height of side PQ =
(i)
8.00 cm
(ii)
14.00 cm
(iii)
11.00 cm
(iv)
6.00 cm
(v)
16.00 cm
Question
13
13.
In a right angled triangle △PQR, if QR = 11 cm, PQ = 13 cm are the lengths of perpendicular sides , then corresponding height of side PQ =
(i)
11.00 cm
(ii)
16.00 cm
(iii)
6.00 cm
(iv)
8.00 cm
(v)
14.00 cm
Question
14
14.
In an isosceles right angled triangle △PQR, if RP = 24.04 cm is the hypotenuse, then corresponding height of side PQ =
(i)
20.00 cm
(ii)
22.00 cm
(iii)
14.00 cm
(iv)
12.00 cm
(v)
17.00 cm
Question
15
15.
In a right angled triangle △PQR, if the area is 123.5 sq.cm and corresponding height of side QR = 19 cm, then side PQ =
(i)
19.00 cm
(ii)
14.00 cm
(iii)
16.00 cm
(iv)
24.00 cm
(v)
22.00 cm
Question
16
16.
In the given figure, △CDE, CF ⟂ DE. Which of the following are true?
a)
CF
2
=
2
DF
.
EF
b)
CD
2
−
CE
2
=
DF
2
−
EF
2
c)
CD
2
+
DF
2
=
CE
2
+
EF
2
d)
CD
2
−
DF
2
=
CE
2
−
EF
2
e)
CD
2
+
CE
2
=
DF
2
+
EF
2
(i)
{a,b}
(ii)
{c,d}
(iii)
{b,d}
(iv)
{e,a,b}
(v)
{c,d,b}
Question
17
17.
In an isosceles right angled triangle △PQR, if RP = 15.56 cm is the hypotenuse, then perimeter of the triangle =
(i)
32.56 cm
(ii)
37.56 cm
(iii)
34.56 cm
(iv)
42.56 cm
(v)
40.56 cm
Question
18
18.
In a right angled triangle △PQR, if the base QR = 12 cm and the corresponding height is 15 cm, then corresponding height of side RP =
(i)
9.37 cm
(ii)
7.37 cm
(iii)
10.37 cm
(iv)
11.37 cm
(v)
8.37 cm
Question
19
19.
In the given figure, △DEF is a triangle in which DE = DF and G is a point on EF. Then
(i)
DE
2
+
DG
2
=
EF
2
(ii)
DE
2
+
DG
2
=
EG
.
FG
(iii)
DE
2
−
DG
2
=
EG
.
FG
(iv)
DE
2
−
DG
2
=
DG
.
EG
(v)
DE
2
−
DG
2
=
DG
.
FG
Question
20
20.
In the given figure, △BCD is an obtuse angled triangle and BE ⟂ CD. Then
(i)
BD
2
=
BC
2
+
CD
2
−
2
CD
.
CE
(ii)
BD
2
=
BC
2
+
CD
2
+
2
BC
.
CD
(iii)
BD
2
=
BC
2
+
CD
2
+
2
CD
.
CE
(iv)
BD
2
=
BC
2
+
CD
2
+
CE
2
(v)
BD
2
=
BC
2
+
CD
2
+
2
CE
.
DE
Question
21
21.
In a right angled triangle △PQR, if the area is 127.5 sq.cm and corresponding height of side QR = 15 cm, then corresponding height of side PQ =
(i)
20.00 cm
(ii)
17.00 cm
(iii)
12.00 cm
(iv)
14.00 cm
(v)
22.00 cm
Question
22
22.
Two poles of heights 9 m and 14 m stand vertically on a plane ground. If the distance between their feet is 14 m, find the distance between their tops
(i)
16.87 m
(ii)
13.87 m
(iii)
15.87 m
(iv)
12.87 m
(v)
14.87 m
Question
23
23.
In a right angled triangle △PQR, if the base QR = 18 cm and the corresponding height is 19 cm, then corresponding height of side PQ =
(i)
23.00 cm
(ii)
18.00 cm
(iii)
15.00 cm
(iv)
21.00 cm
(v)
13.00 cm
Question
24
24.
In the given figure, in △FGH, 'O' is a point inside the triangle. OI ⟂ GH, OJ ⟂ FH and OK ⟂ FG. Then
(i)
FK
2
+
GI
2
+
HJ
2
=
FG
2
+
IH
2
+
HF
2
−
GK
2
−
HI
2
−
JF
2
(ii)
FK
2
+
GI
2
+
HJ
2
=
OF
2
+
OG
2
+
OH
2
+
OI
2
+
OJ
2
+
OK
2
(iii)
FK
2
+
GI
2
+
HJ
2
=
OK
2
+
OJ
2
+
OI
2
(iv)
FK
2
+
GI
2
+
HJ
2
=
OF
2
+
OG
2
+
OH
2
−
OI
2
−
OJ
2
−
OK
2
Question
25
25.
In the given figure, ABC is a triangle and 'O' is a point inside △ABC. The angular bisector of ∠BOA, ∠COB & ∠AOC meet AB, BC & CA at D, E & F respectively . Then
(i)
AD . BE . CF = AB . BC . CA
(ii)
AD . BE . CF = DE . EF . FD
(iii)
AD . BE . CF = DB . EC . FA
(iv)
AD . BE . CF = OA . OB . OC
(v)
AD . BE . CF = OD . OE . OF
Assignment Key
1) (ii)
2) (ii)
3) (iv)
4) (v)
5) (iii)
6) (iv)
7) (ii)
8) (iv)
9) (v)
10) (ii)
11) (ii)
12) (iii)
13) (i)
14) (v)
15) (i)
16) (iii)
17) (ii)
18) (i)
19) (iii)
20) (iii)
21) (ii)
22) (v)
23) (ii)
24) (iv)
25) (iii)
