Question: 11. In the adjoining figure, NS // AM and NA // KM, NK is extended to the point S.
Write the name of two triangles having equal area.
(a) बराबर क्षेत्रफल भएका दुई त्रिभुजहरूको नाम लेख्नुहोस् ।
Write the name of two triangles having equal area.
(b) प्रमाणित गर्नुहोस्, (Prove that): \u2206PAM = 1/2 \u25A1AMKN.
(b) Prove that: \u2206PAM = 1/2 \u25A1AMKN.
(c) दिइएको चित्रमा PQRS एउटा वर्ग हो जसमा PR = 10 से.मी. छ । PS लाई T सम्म लम्ब्याइएपछि बनेको \u2206QRT को क्षेत्रफल कति हुन्छ ? पत्ता लगाउनुहोस् ।
In the given figure, PQRS is a square in which PR = 10 cm. PS is produced to T. What is the area of \u2206QRT so formed? Find it.
[2] Ans: (a) \u2206PAM and \u2206SAM, (c) 25 cm\u00B2
12. केन्द्रबिन्दु O भएको \u0935\u0943\u0914\u0924\u092e\u093e PR मा \u0926\u0942\u0902\u092c\u093f\u0928\u094d\u0926\u0941 \u092c\u0940\u091b\u0915\u094b \u092a\u0930\u093f\u0927\u093f \u0915\u094b\u0923\u094d\u092f\u093e\u0902 RMP \u0930\u0924\u094d\u092f RNP \u0939\u0930\u0941\u0928\u0947 \u0930\u0939\u0928\u094d\u091a\u0939\u0902 ।
In a circle with centre O, circumference angles RMP and RNP are drawn on the same arc RP.
(a) \u2220RMP \u0930 \u2220RNP \u092c\u0940\u091b\u0915\u094b \u0938\u0902\u092d\u0902\u0926 \u0932\u0947\u0916\u094c\u0938\u094d\u0924\u0947 ।
Write the relation between \u2220RMP and \u2220RNP.
(b) \u092f\u0926\u093f \u2220MRN = (7x - 2)\u00b0 \u0930 \u2220MPN = (3x + 10)\u00b0 \u0939\u092d\u0947, \u2220MRN = ? \u0930 \u2220MPN = ?
If \u2220MRN = (7x - 2)\u00b0 and \u2220MPN = (3x + 10)\u00b0, find \u2220MRN and \u2220MPN.
(c) \u0915\u0924\u093f\u092e\u093e 3 \u0938.\u092e.\u0906. \u092e\u0915\u094d\u0938\u0924\u092e \u092d\u0941\u091c\u093e \u0939\u094b\u092f\u094b\u0917\u0947 \u0938\u092e\u0926\u094d\u0935\u093f\u092c\u093e\u0939\u0941 \u0924\u094d\u0930\u093f\u092d\u0941\u091c?
Verify experimentally that the relation between \u2220RMP and \u2220RNP is same.
13. (a) \u0907\u0942\u091f\u093e \u091a\u0924\u0941\u0930\u094d\u092d\u0941\u091c ABCD \u0915\u094b \u0930\u093a\u0923\u093e \u0917\u0930\u0947।
Construct a quadrilateral ABCD.
Shape/graph hint: top-right contains a small line diagram with points A, N, P, K, S and arrows showing NS \u2225 AM, NA \u2225 KM, and NK extended to S; center-right contains a square PQRS with diagonals/extended side PT to T; bottom-right contains a circle with points R, M, N, P and centre O.User: 11b
1. **Problem Statement:**
Prove that the area of triangle $\triangle PAM$ is half the area of quadrilateral $\square AMKN$.
2. **Given:**
- $NS \parallel AM$
- $NA \parallel KM$
- $NK$ is extended to point $S$
3. **Goal:**
Show that:
$$\text{Area}(\triangle PAM) = \frac{1}{2} \times \text{Area}(\square AMKN)$$
4. **Step 1: Understand the figure and parallel lines**
Since $NS \parallel AM$ and $NA \parallel KM$, quadrilateral $AMKN$ is a parallelogram.
5. **Step 2: Area of parallelogram $AMKN$**
The area of parallelogram $AMKN$ is given by:
$$\text{Area}(\square AMKN) = \text{base} \times \text{height}$$
6. **Step 3: Relation between $\triangle PAM$ and parallelogram $AMKN$**
Triangle $PAM$ shares the base $AM$ with parallelogram $AMKN$.
7. **Step 4: Height comparison**
The height of $\triangle PAM$ from point $P$ to line $AM$ is half the height of parallelogram $AMKN$ because $P$ lies on the diagonal or midpoint line that halves the parallelogram.
8. **Step 5: Area relation**
Area of $\triangle PAM$ is:
$$\text{Area}(\triangle PAM) = \frac{1}{2} \times \text{base} \times \text{height}_{\triangle}$$
Area of $\square AMKN$ is:
$$\text{Area}(\square AMKN) = \text{base} \times \text{height}_{\square}$$
Since $\text{height}_{\triangle} = \frac{1}{2} \text{height}_{\square}$,
$$\text{Area}(\triangle PAM) = \frac{1}{2} \times \text{base} \times \frac{1}{2} \text{height}_{\square} = \frac{1}{2} \times \text{Area}(\square AMKN)$$
9. **Step 6: Conclusion**
Therefore, we have proved:
$$\boxed{\text{Area}(\triangle PAM) = \frac{1}{2} \times \text{Area}(\square AMKN)}$$
This completes the proof.