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Answered on 03 Jul Learn Quadrilaterals

Deepika Agrawal

"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."

Given: In square ABCD, AK = BL = CM = DN.To prove: KLMN is a square. In square ABCD, AB = BC = CD = DA And, AK = BL = CM = DN (All sides of a square are equal.) (Given) So, AB - AK = BC - BL = CD - CM = DA - DN ⇒ KB = CL = DM = AN.......... (1) In △NAKand△KBL∠NAK=∠KBL=900 (Each angle of... read more

Given: In square ABCD, AK = BL = CM = DN.
To prove: KLMN is a square.

In square ABCD,

AB = BC = CD = DA

And, AK = BL = CM = DN

(All sides of a square are equal.) (Given)

So, AB - AK = BC - BL = CD - CM = DA - DN

⇒ KB = CL = DM = AN.......... (1)

In NAKandKBL
NAK=KBL=900 (Each angle of a square is a right angle.)
AK=BL (Given)
AN=KB [From (1)]
So, by SAS congruence criteria,
\(\triangle NAK ≅ \triangle KBL \)
⇒NK=KL (Cpctc ...(2)
Similarly,
\(\triangle MDN≅ \triangle NAK \)

\(\triangle DNM≅ \triangle CML \)

\(\triangle MCL≅ \triangle LBK \)
\(\rightarrow MN = NK and \angle DNM=\angle KNA \) (Cpctc )… 3)
MN = JM and DNM=CML (Cpctc )… 4)
ML = LK and CML=BLK (Cpctc )… (5)
From (2), (3), (4) and (5), we get
NK = KL = MN = ML…….....(6)
And, DNM=AKN=KLB=LMC
Now,
In NAK
NAK=900
Let AKN=x0

So, DNK=900+x0 (Exterior angles equals sum of interior opposite angles.)
DNM+MNK=900+x0

x0+MNK=900+x0

⇒ MNK=900
Similarly,
NKL=KLM=LMN=900 ...(7)
Using (6) and (7), we get
All sides of quadrilateral KLMN are equal and all angles are \ ( 90^0 \)
So, KLMN is a square.

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Answered on 16 Jul Learn Quadrilaterals

Deepika Agrawal

"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."

With the given details, we can create a diagram. Please refer to video for the diagram.From the figure, SM||QR||PS∴∠QRN=∠SMRAlso, PQ||SR||QN∴∠RSM=∠RQNIn ΔRSMandΔNQR,∠QRN=∠SMR∠RSM=∠RQNAs two of their angles are equal, third angle will also be equal.... read more

With the given details, we can create a diagram. Please refer to video for the diagram.
From the figure, SM||QR||PS
∴∠QRN=∠SMR
Also, PQ||SR||QN
∴∠RSM=∠RQN
In ΔRSMandΔNQR,
∠QRN=∠SMR
∠RSM=∠RQN
As two of their angles are equal, third angle will also be equal. So, ΔRSM≅ΔNQR
∴SMQR=SRQN⇒SMSR=QRQN
As, SM=SR , it means ,QR=QN

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Answered on 16 Jul Learn Quadrilaterals

Deepika Agrawal

"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."

R.E.F image Since DE||BC and D is mid point of AB, by mid point theorem E is mid point AC and BC =2DE =10cm ∴ perimeter =3.5+3.5+4.5+4.5+10 P=26cm read more

R.E.F image

Since DE||BC and D is mid point of AB, by mid point theorem E is mid point AC and BC =2DE
=10cm
 perimeter =3.5+3.5+4.5+4.5+10
P=26cm
1234291_1179145_ans_c606c859757c408da5a8a412899dbfdf.JPG
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Answered on 16 Jul Learn Quadrilaterals

Deepika Agrawal

"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."

Let ABCD be a parallelogram.∴∠A=∠C and ∠B=∠D (Opposite angles)Let ∠A=x0 and ∠B=4x50Now, ∠A+∠B=1800(Adjacent angles are supplementary)⇒x+4x50=1800 ⇒9x5=1800 ⇒x=20×5 ⇒x=1000 Now, ∠A=1000 and ∠B=45×1000=800 Hence, ∠A=&ang... read more

Let ABCD be a parallelogram.
A=C and B=D (Opposite angles)
Let A=x0 and B=4x50
Now, A+B=1800
(Adjacent angles are supplementary)
x+4x50=1800

9x5=1800

x=20×5

x=1000

Now, A=1000 and B=45×1000=800

Hence, A=C=1000

B=D=800

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Answered on 03 Jul Learn Construction

Deepika Agrawal

"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."

Steps of construction:1. Draw a line segment PQ = 11.6 cm.2. Construct an angle of 45° and bisect it to get ∠QPX.3. Construct an angle of 60° and bisect it to get ∠PQY.4. The ray XP and YQ intersect at A.5. Draw the right bisectors of AP and AQ, cutting PQ at B and C, respectively.6.... read more

Steps of construction:
1. Draw a line segment PQ = 11.6 cm.
2. Construct an angle of 45° and bisect it to get ∠QPX.
3. Construct an angle of 60° and bisect it to get ∠PQY.
4. The ray XP and YQ intersect at A.
5. Draw the right bisectors of AP and AQ, cutting PQ at B and C, respectively.
6. Join AB and AC.
Thus, △ABC is the required triangle.

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Answered on 03 Jul Learn Construction

Deepika Agrawal

"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."

It is given that AB: BC: CA = 3: 4: 5 3x + 4x + 5x = 12 12x = 12 x = 1 AB = 3 cm, BC = 4 cm and CA = 5 cm Steps of construction: (1) Draw a sufficiently long line segment using a ruler. (2) Locate points A and B on it such that AB = 3 cm. (3) With A as the centre and radius 5 cm, draw an arc. (4) With... read more

It is given that AB: BC: CA = 3: 4: 5

3x + 4x + 5x = 12

12x = 12

x = 1

AB = 3 cm, BC = 4 cm and CA = 5 cm

Steps of construction:

(1) Draw a sufficiently long line segment using a ruler.

(2) Locate points A and B on it such that AB = 3 cm.

(3) With A as the centre and radius 5 cm, draw an arc.

(4) With B as the centre and radius 4 cm, draw another arc that cuts the previous arc at C.

(5) Join AC and BC.

Then, ABC is the required triangle.

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Answered on 11 Aug Learn Construction

Deepika Agrawal

"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."

Solution Steps of constructionStep I Draw a line AB = 3.2 cm.Step II Construct a right angle (90∘) at point B, i.e. ∠ABY=90∘.Step III Now, from point A, cut an arc 5 cm on BY at C.Step IV Joint C to A. Hence, ΔABC is the required triangle, having hypotenuse AC = 5 cm and AB = 3.2 cm. read more
Solution
 

Steps of construction
Step I Draw a line AB = 3.2 cm.
Step II Construct a right angle (90) at point B, i.e. ABY=90.
Step III Now, from point A, cut an arc 5 cm on BY at C.
Step IV Joint C to A.

Hence, ΔABC is the required triangle, having hypotenuse AC = 5 cm and AB = 3.2 cm.

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Answered on 11 Aug Learn Construction

Deepika Agrawal

"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."

Steps of construction:1. Draw a line segment XY.2. Take a point D on XY and draw PD ⊥ XY.3. Along PD, set off DA = 4.8 cm.4. Draw a line LM ∥ XY.5. Draw ∠LAB = 30° and ∠MAC = 60°, meeting XY at B and C, respectively.Thus, ABC is the required triangle. read more

Steps of construction:
1. Draw a line segment XY.
2. Take a point D on XY and draw PD ⊥ XY.
3. Along PD, set off DA = 4.8 cm.
4. Draw a line LM ∥ XY.
5. Draw ∠LAB = 30° and ∠MAC = 60°, meeting XY at B and C, respectively.
Thus, ABC is the required triangle.

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Answered on 11 Aug Learn Construction

Deepika Agrawal

"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."

Construct an equilateral triangle whose perimeter is 16.2 cm Draw a line BC = 5.4 cm. Using a compass, take radius 5.4 cm and using B as a centre draw an arc. Taking the same radius and using C as a centre, cut the previous arc at A. Join AB and AC. read more
Construct an equilateral triangle whose perimeter is 16.2 cm
  1. Draw a line BC = 5.4 cm.
  2. Using a compass, take radius 5.4 cm and using B as a centre draw an arc.
  3. Taking the same radius and using C as a centre, cut the previous arc at A.
  4. Join AB and AC.

construct an equilateral triangle whose perimeter is 16.2 cm ...

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Answered on 11 Aug Learn Construction

Deepika Agrawal

"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."

The correct option is A 6.9cmGiven, BC = 6cm and ∠B=45∘We know that, the construction of a triangle is not possible, if sum of two sides is less than or equal to the third side of the triangle.i.e., AB+BC≤AC⇒ BC≤AC−AB⇒ 6≤AC−ABSo, if AC... read more

The correct option is A 6.9cm
Given, BC = 6cm and B=45
We know that, the construction of a triangle is not possible, if sum of two sides is less than or equal to the third side of the triangle.
i.e., AB+BCAC
BCACAB
6ACAB
So, if AC – AB = 6.9cm, then construction of ΔABC with given conditions is not possible.

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