3. Verify whether the following are zeroes of the polynomial, indicated against them.
1
4
(ii) p(x) = 5x - , *
5
() p(x)= 3x + 1, x= --
3
(iii) p(x)=x2-1, x= 1,-1
(iv) p(x)=(x + 1)(x - 2), x=-1,2
m
(v) p(x)= x², x=0
(vi) p(x)=1x + m, x= --
1
1 2
1
(viii) p(x)=2x + 1, x=
2
x=
(vii) p(x)= 3x2 - 1, x= -
53 53
Answers
Answer:
DEF are respectively the midpoints of sides AB, BC and CA of triangle ABC, then: Ratio of area of triangle DEF : area of triangle ABC = 1 : 4
Solution:
In the given question, we know Triangle DEF formed with midpoints is similar to the Outer Triangle ABC
On the basis of the similarity, we can say,
If two triangles are similar then the ratio of their area is equal to the square of the ratio of their corresponding sides
Mathematically can be written as :-
\frac{\text {area of triangle } D E F}{\text {area of triangle ABC}}=\left(\frac{D E}{A C}\right)^{2}
area of triangle ABC
area of triangle DEF
=(
AC
DE
)
2
Since, DECF is a parallelogram. So DE = FC
On substituting:
\frac{\text {area of triangle } D E F}{\text {area of triangle ABC}}=\left(\frac{F C}{A C}\right)^{2}
area of triangle ABC
area of triangle DEF
=(
AC
FC
)
2
Also, F is the midpoint of AC
\text { So } \mathrm{AC}=2 \times \mathrm{FC} So AC=2×FC
\begin{gathered}\begin{array}{l}{\frac{\text {area of triangle } D E F}{\text {area of triangle } A B C}=\left(\frac{F C}{2 F C}\right)^{2}} \\\\ {\frac{\text {area of triangle } D E F}{\text {area of triangle } A B C}=\frac{1}{4}}\end{array}\end{gathered}
area of triangle ABC
area of triangle DEF
=(
2FC
FC
)
2
area of triangle ABC
area of triangle DEF
=
4
1
Hence ratio of area of triangle DEF and triangle ABC is given as:
Ratio of area of triangle DEF : area of triangle ABC = 1 : 4
Learn more about midpoints of triangle
Find the vertices of a triangle, the midpoints of whose side are(3,1),(5,6,),(-3,2)
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The mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to what?
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Answer:
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Step-by-step explanation:
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