Take Class 10 Tuition from the Best Tutors
Affordable fees 1-1 or Group class Flexible Timings Verified Tutors Search in
Ask a Question
Post a LessonAnswered 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 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.
Answered on 16 Jul Learn Quadrilaterals
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Answered on 16 Jul Learn Quadrilaterals
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Take Class 10 Tuition from the Best Tutors
Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors 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=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
read lessAnswered 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. Join AB and AC.
Thus, △ABC is the required triangle.
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 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.
Take Class 10 Tuition from the Best Tutors
Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors 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
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.
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.
Answered on 11 Aug Learn Construction
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Take Class 10 Tuition from the Best Tutors
Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors 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 6.9cm
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+BC≤AC
⇒ BC≤AC−AB
⇒ 6≤AC−AB
So, if AC – AB = 6.9cm, then construction of ΔABC with given conditions is not possible.
UrbanPro.com helps you to connect with the best Class 10 Tuition in India. Post Your Requirement today and get connected.
Ask a Question
The best tutors for Class 10 Tuition Classes are on UrbanPro
The best Tutors for Class 10 Tuition Classes are on UrbanPro
124,210 
