6. Triangles
Triangles
6. TRIANGLES
SECTION = A
Q1. If 
and ar(
) = 
ar(
), then the value of 
is :
(a) 
(b) 
(c) 
(d)
Solution: (c)
[ Since, 
DEF 
ABC .
Given , 
]
Q2. In given figure , S and T are points on the sides PQ and PR , respectively of ∆PQR , such that PT = 2 cm and TR = 4 cm and ST is parallel to QR ,then the ratio of the area of ∆PST and ∆PQR :
(a) 
(b) 
(c) 
(d) 
Solution: (b)
[ Since ST
QR , then 
.
A/Q , 
]
Q3. In figure , DEBC ,then EC is equal to :
(a) 2 cm (b) 3 cm (c) 5 cm (d) 6 cm
Solution: (a) 2 cm [ In 
and 
we have ,
cm ]
Q4. Which of the following given the sides of the triangle make is a right triangle ?
(a) 3 cm , 8 cm , 6 cm (b) 50 cm , 80 cm , 100 cm
(c) 25 cm , 24 cm , 7 cm (d) 7 cm , 11 cm , 13 cm
Solution: (c) 25 cm , 24 cm , 7 cm
[ Here , 
cm , 
cm and 
cm
Therefore , 
is a right triangle . ]
Q5. ABC and BDE are two equilateral triangles such that D is the mid-point of BC .
Ratio of the areas of triangles ABC and BDE is :
(a) 2 : 1 (b) 1 : 2 (c) 4 : 1 (d) 1 : 4
Solution: (c) 4 : 1
[ Since ABC and BDE ae two equilitarel triangles , i.e., ABC 
BDE .
So, 
4 : 1 ]
Q6. If 
and 
, then RQ is :
(a) 6 cm (b) 12 cm (c) 10 cm (d) 3 cm
Solution: (b) 12 cm
[ Since 
, we have
So, 
]
Q7. Let ABC be a triangle such that AB = 
cm , AC = 12 cm and BC = 6 cm ,
then 
is :
(a) 120° (b) 60° (c) 90° (d) 45°
Solution: (c) 90°
[ InABC , we have 
.
So, 
]
Q8. In given figure , MNAB , AB = 7.5 cm , AM = 4 cm and MC = 2 cm , then the length of BN is :
(a) 5 cm (b) 4 cm (c) 2 cm (d) 8 cm
Solution: (a) 5 cm
[ In ∆ABC and MNAB , then ABCMNC ,
we have 
cm 
cm ]
Q9. In DEW , ABEW . If 
and 
,then the value of DB is :
(a) 12 cm (b) 24 cm (c) 8 cm (d) 4 cm
Solution: (c) 8 cm
[ InDEW and ABEW,
We have, 
cm ]
Q10. DE is drawn parallel to the base BC of a ABC , meeting AB at D and AC at E . If 
and CE = 2 cm , then AE is :
(a) 5 cm (b) 4 cm (c) 6 cm (d) 7 cm
Solution: (c) 6 cm
[ In ∆ABC and DEBC,
We have , 
.
So, 
cm ]
Fill in the blank
Q1. All circles are 
. [ congruent / similar]
Solution: Similar .
Q2. All squares are 
. [ similar / congruent ]
Solution: Similar .
Q3. All triangles are similar . [ isosceles / equilateral / acute triangle ]
Solution: Equilateral .
Q4. Two polygons of the same number of sides are similar , if
(a) their corresponding angles are
and (b) their corresponding sides are 
.
[congruent / equal / proportional /Similar ]
Solution: Equal , Proportional .
Q5. is an isosceles triangle in which 
90° . If AC= 6 cm , then .
Solution: 
cm
[ In ,we have
[ ∵
BC = AC ]
cm ]
SECTION = B
Q1. ABC is an equilateral triangle of side 2a . Find each of its altitudes .
Solution: Since ABC be an equilateral triangle AB = BC = AC = 
We draw AD
BC then
In 
, we have 
Q2. E is a point on side CB produced of an isosceles triangle ABC with 
. If 
and 
, prove that 
.
Solution: Given, E is a point on side CB produced of an isosceles triangle ABC with 
and and .
To prove : 
.
Proof : In ABC , we have
i. e. 
In 
and 
, we have
[ Given]
[ A.A rule ] Proved .
Q3. In given figure, if 
and 
, prove that 
.
Solution: In 
and we have ,
Again, 
and we have ,
and 
we have , 
Proved.
Q4. In given figure , If , prove that 
.
Solution: In 
, we have
In , we have
Proved .
SECTION = C
Q1. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C . Prove that 
.
Solution: Given, 
and
are points on the sides
and
respectively of a triangle right angled at C .
To Proved : 
Proof : In we have,
In 
we have,
In 
we have,
In 
we have,
and 
we get ,
[ From and 
]
Proved .
Q2. In figure , ABC and DBC are two triangles on the same base BC . If AD intersects BC at O , show that 
.
Solution: Given, and 
are two triangles on the same base 
and 
intersects at O .
To Prove : 
Construction : We draw 
and 
.
Proof : In 
and 
We have ,
[ Vertical opposite angle]
[ Alternative interior angle]
[A-A-A rule]
So , 
……………… (i)
[ From (i) ] Proved .
SECTION = D
1. ABCD is a trapezium with AB∥DC . E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB . Show that 
.
Solution: Given, ABCD is a trapezium with ABDC . E and F are points on non-parallel sides AD and BC respectively such that EF AB .
To Prove :
Construction : Join AC to intersect EF at G .
Proof : ABDC and EFAB and Also EFDC
In ADC and EGDC ( EFDC )
So, 
………….. (i)
In ACB and GFAB ( GFAB )
So, 
………….. (ii)
From (i) and (ii) , we get
Proved.
2. In a triangle, if square of one side is equal to the sum of the squares of the other two sides , then the angle opposite the first side is a right angle .
Solution: Given , ABC be a triangle in which 
.
To Prove :
Construction : We draw a 
right angled at Q such that PQ = AB and QR = BC .
Proof : In , we have
[ 
]
[ PQ = AB and QR = BC]
Again , 
From and 
, we get 
In and , we have
[ S.S.S congruence]
[ CPCT]
Proved .
3. In figure, the line segment XY is parallel to side AC of ∆ABC and it divides the triangle into two parts of equal areas . Find the ratio 
.
Solution: Since, the line segment XY is parallel to side AB of the triangle ABC .
Therefore, 
and 
[ corresponding angles]
In 
and , we have
[ Given]
[ Given]
[ A.A rule]
So, 
Given, 
From and , we get 
Q4. The perpendicular from A on side BC of a ABC intersects BC at D such that 
. Prove that 
.
Solution: Given, be a triangle and and intersect at 
such that .
To prove :
Proof : Given, 
and 
In we have ,
and we have ,
Proved.
5. In a right triangle , the square of the hypotenuse is equal to the sum of the squares of the other two sides .
Solution: Given , ABC be a right triangle and 
B = 90° .
To prove : .
Construction : we draw BD
AC .
Proof : In and , we have
[ Common angle]
[ A. A. ]
So, 
In 
and , we have
[ Common angle]
[ A. A. ]
So, 
Adding and , we get 
Proved .
6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides .
Solution: Given , ABCD be a parallelogram and diagonals AC and BD intersecting at a point O .
To Prove : 
.
Proof : we know that diagonals of a parallelogram bisect each other .
i.e. , 
and 
Since OB be a median of ABC , then
…………… (i)
Again, OD be a median of ADC , then
…………… (ii)
Adding (i) and (ii) , we get 
Proved .
SECTION = E
Q1. Prove that a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio .
Solution: Given, a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively .
To prove : .
Construction : Join BE and CD and also , draw DMAC and ENAB .
Proof : We know that , Area of triangle 
× Base × Height
and 
and 
Since 
and 
are on the same base DE and between the same parallels BC and DE.
So, 
From 
, 
and 
, we have
Proved.
Q2. O is any point inside a rectangle ABCD . Prove that 
.
Solution: Given, 
is any point inside a rectangle ABCD .
To Prove : .
Construction : Through O, draw PQBC and P lies on AB and Q lies on DC .
Proof : Since,
and 
.
and 
Therefore , 
and 
are both rectangles .
In 
we have ,
In 
we have ,
In 
we have ,
In 
we have ,
[
,
]
Proved .
Q3. In figure , O is a point in the interior of a triangle ABC , 
, 
and 
. Show that 
Solution: Given O is a point in the interior of a triangle ABC , , and .
To Prove :
Construction : We join 
,
and 
Proof : In figure,
In 
we have ,
In 
we have ,
In 
we have ,
In 
we have ,
In 
we have ,
In 
we have ,
From 
and 
, we have
Proved.
Q4. Prove the ratio the areas of two similar triangles is equal to the square of the ratio of their corresponding sides . [CBSE 2020 standard]
Solution: Given, ABC and PQR are two triangle such that 
.
To Prove : 
Construction : We draw AMBC and PNQR .
Proof : In ABC , we have
In PQR , we have
In ABM and PQN , we have
[ ]
In 
[ A.A ]
In [ Given ]
[ From 
and 
]
Similarly , we show that 
Proved.
Posted 5 years ago