10th Maths 7.3
NCERT Class 10th solution of Exercise 7.1
Exercise 7.3
Q1. Find the area of the triangle whose vertices are:
i) (2,3),(-1,0),(2,-4)
ii) (-5,-1),(3,-5),(5,2)
Sol. :
i)
Given:
Let A(2,3),B(-1,0), and C(2,-4)
Solve:
Area of △=12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
ar(ABC)=12[2(0+4)+(-1)(-4-3)+2(3-0)]
ar(ABC)=12[2(4)+(-1)(-7)+2(3)]
ar(ABC)=12[8+7+6]
ar(ABC)=212 square units
Answer:
The area of △ABC is 212 square units.
ii)
Given:
Let A(-5,-1),B(3,-5), and C(5,2)
Solve:
Area of △=12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
ar(ABC)=12[-5(-5-2)+3(2+1)+5(-1+5)]
ar(ABC)=12[-5(-7)+3(3)+5(4)]
ar(ABC)=12[35+9+20]
ar(ABC)=12[64]
ar(ABC)=32 square units
Answer:
The area of △ABC is 32 square units.
Q2. In each of the following find the value of k, for which the points are collinear.
i) (7,-2),(5,1),(3,k)
ii) (8,1),(k,-4),(2,-5)
Sol. :
i)
Given:
Let A(7,-2),B(5,1), and C(3,k) are collinear.
Solve:
We know that
The points A,B, and C are collinear, then ar(ABC)=0.
Area of △=12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
ar(ABC)=12[7(1-k)+5(k+2)+3(-2-1)]
0=12[7-7k+5k+10+3(-3)]
0=12[17-2k-9]
0=12[8-2k]
0=8-2k
2k=8
k=82
k=4
Answer:
The value of k=4.
ii)
Given:
Let A(8,1),B(k,-4), and C(2,-5) are collinear.
Solve:
We know that
The points A,B, and C are collinear then ar(ABC)=0.
Area of △=12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
ar(ABC)=12[8(-4+5)+k(-5-1)+2(1+4)]
0=12[8(1)+k(-6)+2(5)]
0=12[8-6k+10]
0=12[18-6k]
k=186
`k=3
Answer:
The value of k=3.
Q3. Find the area of the triangle formed by joining the mid-points to the sides of the triangle whose vertices are (0,-1),(2,1) and (0,3). Find the ratio of this area to the area of the given triangle.
Sol. :
Let A(0,-1),(2,1), and C(0,3)
To Find:
The ratio of the formed △ area of the given △.
Solve:
The midpoint of AB is P
P=(0+22,-1+12)=(1,0)
The midpoint of BC is Q
Q=(2+02,1+32)=(1,2)
The midpoint of CA is R
R=(0+02,3-12)=(0,1)
Area of △=12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
ar(ABC)=12[0(1-3)+2(3+1)+0(-1-1)]
ar(ABC)=12[0+2(4)+0]
ar(ABC)=12×8=4 square units
And
ar(PQR)=12[1(2-1)+1(1-0)+0(0-2)]
ar(PQR)=12[1+1-0]
ar(PQR)=12×2=1 square units
The ratio of ar(ABC) and ar(PQR)
ar(ABC)ar(PQR)=14
ar(ABC):ar(PQR)=1:4
Answer:
The ratio of the formed △ area to the given △ is 1:4.
Q4. Find the area of the quadrilateral whose vertices, takes in order, are (-4,-2),(-3,-5),(3,-2) and (2,3).
Sol. :
Given:
Let the vertices of parallelogram A(-4,-2),B(-3,-5),C(3,-2) and D(2,3)
Construction:
Join AC
To Find:
The area of the parallelogram ABCD.
Solve:
Area of △=12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
ar(ABC)=12[-4(-5+2)+(-3)(-2+2)+3(-2+5)]
ar(ABC)=12[-4(-3)+(-3)(0)+3(3)]
ar(ABC)=12[12-0+9]
ar(ABC)=12×21=212 square units
and
ar(ACD)=12[-4(-2-3)+3(3+2)+2(-2+2)]
ar(ACD)=12[-4(-5)+3(5)+2(0)]
ar(ACD)=12[20+15+0]
ar(ACD)=12[35]=352 square units
ar(ABCD)=ar(ABC)+ar(ACD)
ar(ABCD)=212+352=562=28 square units.
Answer:
The area of parallelogram ABCD=28 square units.
Q5. You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal area. Verify this result for △ABC whose vertices are A(4,-6),B(3,-2) and C(5,2).
Sol. :
A(4,-6),B(3,-2) and C(5,2)
To Verify:
A median of a △ divides it into two △s of equal areas.
Solve:
In △ABC AD is median that bisect BC at D
Let D(x,y)
x=3+52=4,
y=-2+22=0
In △ADB
A(4,6),D(4,0),B(3,-2)
Area of △=12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
ar(ADB)=12[4(0+2)+4(-2+6)+3(-6-0)]
ar(ADB)=12[8+16-18]
ar(ADB)=12×6=3 square units
and
In △ADC
A(4,6),B(4,0),C(5,2)
ar(ADC)=12[4(0-2)+4(2+6)+5(-6-0)]
ar(ADC)=12(-8+32-30)
ar(ADC)=12×(-6)=-3 square units.
Since area is a measure, which cannot be negative.
ar(ADC)=3 square units
ar(ADB)=ar(ADC)
A median of the △ divides it into two △s of equal areas.
Verified.
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