Monday, February 8, 2021

Computer test

Computer Test

Computer Test

 To give test, Click on:

Computer TEST

टेस्ट देने के लिए, इस पर क्लिक करें:




Computer Test 

Hello everyone. This is Mohit Sharma. I want to inform you that this is a practice test paper of Computer. First read all the guidelines then after that you can attempt this short quiz. Here a link is given to you at last. Click on the link and attempt this short quiz.

सभी को नमस्कार। मेरा नाम मोहित शर्मा है । मैं आपको सूचित करना चाहता हूं कि यह COMPUTER   का एक अभ्यास परीक्षण पेपर है। सभी दिशानिर्देशों को पढ़ें इसके बाद आप इस संक्षिप्त प्रश्नोत्तरी का प्रयास कर सकते हैं। यहां पर आपको आखिरी में एक लिंक दिया गया है। लिंक पर क्लिक करें और इस संक्षिप्त प्रश्नोत्तरी का प्रयास करें।
ALL THE BEST. ✌

Guidelines:
This is a quiz test. You have to attempt all questions.
Total Questions: 10
Max. marks: 10 (Each question carry 1 mark)
Subject: Computer
School: J.P. PUBLIC ACADEMY, KASGANJ

After attempting short quiz you will get your SCORE on screen with correct answers if you have done any question incorrect.

Friday, February 5, 2021

Computer Test of Class VIII

Computer Test of Class VIII

Computer Test of Class VIII

To give test, Click on:
Computer TEST

टेस्ट देने के लिए, इस पर क्लिक करें:


Computer Test of Class VIII 

Hello everyone. This is Mohit Sharma. I want to inform you that this is a practice test paper of Computer. First read all the guidelines then after that you can attempt this short quiz. Here a link is given to you at last. Click on the link and attempt this short quiz.

सभी को नमस्कार। मेरा नाम मोहित शर्मा है । मैं आपको सूचित करना चाहता हूं कि यह COMPUTER   का एक अभ्यास परीक्षण पेपर है। सभी दिशानिर्देशों को पढ़ें इसके बाद आप इस संक्षिप्त प्रश्नोत्तरी का प्रयास कर सकते हैं। यहां पर आपको आखिरी में एक लिंक दिया गया है। लिंक पर क्लिक करें और इस संक्षिप्त प्रश्नोत्तरी का प्रयास करें।
ALL THE BEST. ✌

Guidelines:
This is a quiz test. You have to attempt all questions.
Total Questions: 10
Max. marks: 10 (Each question carry 1 mark)
Subject: Computer
School: J.P. PUBLIC ACADEMY, KASGANJ

After attempting short quiz you will get your SCORE on screen with correct answers if you have done any question incorrect.

https://amzn.to/3oX8OWZ

Computer Test of Class VI 2nd

Computer Test of Class VIII

Monday, January 25, 2021

NCERT Class 9 Maths Chapter 8 – Quadrilaterals

NCERT Class 9 Maths Chapter 8 – Quadrilaterals

NCERT Class 9 Maths Chapter 8 – Quadrilaterals

 

Exercise 8.2 Page: 150

1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that:
(i) SR || AC and SR = 1/2 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Ncert solutions class 9 chapter 8-12

Solution:

(i) In ΔDAC,

R is the mid point of DC and S is the mid point of DA.

Thus by mid point theorem, SR || AC and SR = ½ AC

(ii) In ΔBAC,

P is the mid point of AB and Q is the mid point of BC.

Thus by mid point theorem, PQ || AC and PQ = ½ AC

also, SR = ½ AC

, PQ = SR

(iii) SR || AC ———————- from question (i)

and, PQ || AC ———————- from question (ii)

⇒ SR || PQ – from (i) and (ii)

also, PQ = SR

, PQRS is a parallelogram.

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Solution:

Ncert solutions class 9 chapter 8-13

Given in the question,

ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

To Prove,

PQRS is a rectangle.

Construction,

Join AC and BD.

Proof:

In ΔDRS and ΔBPQ,

DS = BQ (Halves of the opposite sides of the rhombus)

∠SDR = ∠QBP (Opposite angles of the rhombus)

DR = BP (Halves of the opposite sides of the rhombus)

, ΔDRS ≅ ΔBPQ [SAS congruency]

RS = PQ [CPCT]———————- (i)

In ΔQCR and ΔSAP,

RC = PA (Halves of the opposite sides of the rhombus)

∠RCQ = ∠PAS (Opposite angles of the rhombus)

CQ = AS (Halves of the opposite sides of the rhombus)

, ΔQCR ≅ ΔSAP [SAS congruency]

RQ = SP [CPCT]———————- (ii)

Now,

In ΔCDB,

R and Q are the mid points of CD and BC respectively.

⇒ QR || BD

also,

P and S are the mid points of AD and AB respectively.

⇒ PS || BD

⇒ QR || PS

, PQRS is a parallelogram.

also, ∠PQR = 90°

Now,

In PQRS,

RS = PQ and RQ = SP from (i) and (ii)

∠Q = 90°

, PQRS is a rectangle.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Solution:

Ncert solutions class 9 chapter 8-14

Given in the question,

ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.

Construction,

Join AC and BD.

To Prove,

PQRS is a rhombus.

Proof:

In ΔABC

P and Q are the mid-points of AB and BC respectively

, PQ || AC and PQ = ½ AC (Midpoint theorem) — (i)

In ΔADC,

SR || AC and SR = ½ AC (Midpoint theorem) — (ii)

So, PQ || SR and PQ = SR

As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.

, PS || QR and PS = QR (Opposite sides of parallelogram) — (iii)

Now,

In ΔBCD,

Q and R are mid points of side BC and CD respectively.

, QR || BD and QR = ½ BD (Midpoint theorem) — (iv)

AC = BD (Diagonals of a rectangle are equal) — (v)

From equations (i), (ii), (iii), (iv) and (v),

PQ = QR = SR = PS

So, PQRS is a rhombus.

Hence Proved

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC.

Ncert solutions class 9 chapter 8-15

Solution:

Given that,

ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.

To prove,

F is the mid-point of BC.

Proof,

BD intersected EF at G.

In ΔBAD,

E is the mid point of AD and also EG || AB.

Thus, G is the mid point of BD (Converse of mid point theorem)

Now,

In ΔBDC,

G is the mid point of BD and also GF || AB || DC.

Thus, F is the mid point of BC (Converse of mid point theorem)

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD.

Ncert solutions class 9 chapter 8-16

Solution:

Given that,

ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively.

To show,

AF and EC trisect the diagonal BD.

Proof,

ABCD is a parallelogram

, AB || CD

also, AE || FC

Now,

AB = CD (Opposite sides of parallelogram ABCD)

⇒½ AB = ½ CD

⇒ AE = FC (E and F are midpoints of side AB and CD)

AECF is a parallelogram (AE and CF are parallel and equal to each other)

AF || EC (Opposite sides of a parallelogram)

Now,

In ΔDQC,

F is mid point of side DC and FP || CQ (as AF || EC).

P is the mid-point of DQ (Converse of mid-point theorem)

⇒ DP = PQ — (i)

Similarly,

In ΔAPB,

E is midpoint of side AB and EQ || AP (as AF || EC).

Q is the mid-point of PB (Converse of mid-point theorem)

⇒ PQ = QB — (ii)

From equations (i) and (i),

DP = PQ = BQ

Hence, the line segments AF and EC trisect the diagonal BD.

Hence Proved.

6Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Solution:

Ncert solutions class 9 chapter 8-17

Let ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively.

Now,

In ΔACD,

R and S are the mid points of CD and DA respectively.

, SR || AC.

Similarly we can show that,

PQ || AC,

PS || BD and

QR || BD

, PQRS is parallelogram.

PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.

7. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) CM = MA = ½ AB

Solution:

Ncert solutions class 9 chapter 8-18

(i) In ΔACB,

M is the midpoint of AB and MD || BC

, D is the midpoint of AC (Converse of mid point theorem)

(ii) ∠ACB = ∠ADM (Corresponding angles)

also, ∠ACB = 90°

, ∠ADM = 90° and MD ⊥ AC

(iii) In ΔAMD and ΔCMD,

AD = CD (D is the midpoint of side AC)

∠ADM = ∠CDM (Each 90°)

DM = DM (common)

, ΔAMD ≅ ΔCMD [SAS congruency]

AM = CM [CPCT]

also, AM = ½ AB (M is midpoint of AB)

Hence, CM = MA = ½ AB

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