Question

Kerala is a state in Southern India. The state is known as a tropical paradise of waving

palms and wide sandy beaches.

This map the Indian province of Kerala shows its area can be approximated using a

simple straight – sided shape. The shape has two parallel sides 561 km and 216 km

long. The other sides are 180 km and 211 km long. Its parallel sides are 100 km apart.

Shreya observed the shape formed by four straight lines and explored it on her

notebook in different ways shown below.

Shape I Shape II

Refer to Shape I

a) Let ABCD is a trapezium with AB// DC , E and F are points on non – parallel sides AD and BC

respectively such that EF // AB. Then



=

i)



ii)



iii)



iv) none of these

b) Here, AB//CD. If DO = 3x−1, OB= 5x – 3, AO = 6x – 5 and OC= 2x + 1 , then the value of x is

i) 0 ii) 1 iii) 2 iv) 3

c) Here, AB//CD. If DO = 3x−19, OB= x – 5, AO = 3 and OC= x − 3 , then the value of x is

i) 5 or 8 ii) 8 or 9 iii) 10 or 12 iv) none of these

Refer to Shape II

d) In ∆ ABC , PQ//BC . If AP=2.4 cm , AQ =2 cm , QC = 3 cm and BC = 6 cm , AB and PQ are respectively

i) AB = 6 cm , PQ = 2.4 cm ii) AB = 4.8 cm , PQ = 8.2 cm

iii) AB = 4 cm , PQ = 5.3 cm iv) AB = 8.4 cm , PQ = 2.8 cm

e) In ∆ DEF , if RS// EF , DR = 4x – 3 ,DS= 8x – 7, ER = 3x −1 and FS =5x – 3 , then the value of x is

i) 1 ii) 2 iii) 8 iv) 10

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Answer 1

Answer:

A. (iii)BC/FC

B. (iii)2

C. (ii) 8 or 9

D. (i) AB= 6 cm , PQ= 2.4 cm

E. (i) 1

Answer 2

In the trapezium ABCD, $\frac{AE}{ED}= \frac{BF}{FC}$.

Value of x here, will be 2 (iii).

Value of x here, will be 8 or 9 (ii).

In Δ ABC, AB = 6 cm and PQ = 2.4 cm (i).

In ΔDEF , the value of x will be 1 (i).

Explanation 1

We have been given a trapezium with $AB||DC$ and $EF||AB$ . If we draw a diagonal of a trapezium from B to D with $BD$ intersecting $EF$ at $O$, we get two triangles Δ$DAB$ and Δ$DCB$ .

We know,

• Basic proportionality theorem or Thales's theorem states that in a triangle, if a line is running parallel to any one side of a triangle such that it intersects the other two sides of the triangle, then the two sides are divided in the same ratio.

Applying Basic proportionality theorem in Δ$DAB$, we get

$\frac{AE}{ED}= \frac{BO}{OD}$      $(i)$

Similarly, applying $BPT$ theorem in Δ$DCB$, we get

$\frac{BO}{OD}= \frac{BF}{FC}$       $(ii)$

From $(i)$ and $(ii)$, we get

$\frac{AE}{ED} =\frac{BF}{FC}$

Explanation 2

We have been given that $AB||CD$ ; and

$DO=3x-1\\OB=5x-3\\AO=6x-5\\OC=2x+1$

Since, $AB||CD$, hence we know that

$AO+OD=BO+OC$ as these are the diagonals of the same trapezium.

$6x-5+3x-1=5x-3+2x+1$

$2x=4$

Hence, $x=2$

Explanation 3

Given,

$DO=3x-19\\OB=x-5\\AO=3\\OC=x-3$

We know,

$\frac{AO}{CO}= \frac{BO}{DO}$

$\frac{3}{x-3}=\frac{x-5}{3x-19}$

$3(3x-19)=(x-5)(x-3)\\9x-57=x^{2} -8x+15\\x^{2} -17x+72=0$

Solving this, we get $x=8,9$

Explanation 4

We have been given a Δ$ABC$ with $PQ||BC$ and

$AP=2.4cm\\AQ=2cm\\QC=3cm\\BC=6cm$

Applying Basic proportionality theorem in Δ$ABC$, we get

$\frac{AP}{PB}= \frac{AQ}{QC}$

Substituting the given values in the given equation, we get

$\frac{2.4}{PB}= \frac{2}{3}$    $;$  $PB=3.6 cm$

Hence, $AB=AP+PB=2.4+3.6=6 cm$

Similarly,

Applying, Basic proportionality theorem, we get

$\frac{AP}{AB}= \frac{PQ}{BC}$

$\frac{2.4}{6} =\frac{PQ}{6}$   $;$  $PQ=2.4$

Hence, $PQ=2.4cm$

Explanation 5

We have been given a Δ$FED$ with $RS||EF$ and

$DR=4x-3\\DS=8x-7\\ER=3x-1\\FS=5x-3$

Applying Basic proportionality theorem in Δ$FED$, we get

$\frac{DR}{RE}= \frac{DS}{SF}$

Substituting the given values, we get

$\frac{4x-3}{3x-1}=\frac{8x-7}{5x-3}$

$(4x-3)(5x-3)=(8x-7)(3x-1)\\4x^{2} -2x-2=0$

Solving this, we get $x=-\frac{1}{2},1$

Hence, $x=1$

Although your question is incomplete, you might be referring to the diagram below