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Class VI
Class VII
Class VIII
Class IX
Class X
CBSE Grade X - Triangles
Question
1
1.
In the given figure
△FGH
,
I
is the mid-point of
FG
and
IJ
∥
GH
,
then
FJ
=
FG
2
FI
GH
GH
2
HF
2
Question
2
2.
In the given figure, △CDE is a triangle with CF being the median of DE. Then
CD
2
+
CE
2
=
DE
2
CD
2
+
CE
2
=
2
FE
2
+
2
CF
2
CD
2
+
CE
2
=
CF
2
CD
2
+
CE
2
=
2
DF
2
+
2
FE
2
CD
2
+
CE
2
=
2
DF
2
+
2
CF
2
Question
3
3.
In the given figure, given ∠IFG = ∠HFI, p = 7.97 cm, q = 9.03 cm and GH = 17 cm, find IH =
11.03 cm
9.03 cm
7.03 cm
8.03 cm
10.03 cm
Question
4
4.
In the given figure, the parallelogram JKLM and the triangle △NJK are on the same bases and between the same parallels.
The area of the
△NJK
is x sq.cm. The area of the parallelogram is
twice
the area of the triangle
5
4
the area of the triangle
thrice
the area of the triangle
3
2
the area of the triangle
4
3
the area of the triangle
Question
5
5.
In the given figure, three lines l , m and n are such that l ∥ m ∥ n.
Two transversals PQ and RS intersect them at the points A , B , C and D , E , F respectively.
∠DAF =
∠HFE
∠EHF
∠BHA
∠CFA
∠AFD
Question
6
6.
In the given figure, given ∠DAB = ∠CAD, x : y = 8.23 cm : 8.77 cm and q = 16 cm, find p =
15.00 cm
17.00 cm
14.00 cm
16.00 cm
13.00 cm
Question
7
7.
In the given figure, △CDE is isosceles with CD = CE and DF ⟂ CE. Then
DF
2
−
CF
2
= 2
EF
.
CF
DF
2
−
EF
2
= 2
EF
.
CF
DF
2
+
EF
2
= 2
EF
.
CF
DF
2
+
CF
2
= 2
EF
.
CF
Question
8
8.
In the given figure, in △ABC, 'O' is a point inside the triangle. OD ⟂ BC, OE ⟂ AC and OF ⟂ AB. Then
AF
2
+
BD
2
+
CE
2
=
AE
2
+
CD
2
+
BF
2
AF
2
+
BD
2
+
CE
2
=
OD
2
+
OE
2
+
OF
2
AF
2
+
BD
2
+
CE
2
=
OA
.
OB
+
OB
.
OC
+
OC
.
OA
AF
2
+
BD
2
+
CE
2
=
OF
.
OD
+
OD
.
OE
+
OE
.
OF
Question
9
9.
In the given figure, ∠FCD = 42.93°, find the value of x =
48.07°
45.07°
46.07°
47.07°
49.07°
Question
10
10.
In the given figure, BCDE is a rhombus. Which of the following are true?
a)
2
BC
2
=
BD
2
+
CE
2
b)
4
BC
2
=
BD
2
+
CE
2
c)
BC
2
+
CD
2
+
DE
2
+
BE
2
=
BD
2
+
CE
2
d)
CD
2
+
DE
2
=
CE
2
e)
BC
2
+
CD
2
=
BD
2
{a,b}
{b,c}
{d,c,b}
{d,c}
{e,a,b}
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